{"id":307,"date":"2023-07-06T14:13:39","date_gmt":"2023-07-06T14:13:39","guid":{"rendered":"https:\/\/mathority.org\/it\/esempi-di-regole-ed-esercizi-risolti-di-cramer\/"},"modified":"2023-07-06T14:13:39","modified_gmt":"2023-07-06T14:13:39","slug":"esempi-di-regole-ed-esercizi-risolti-di-cramer","status":"publish","type":"post","link":"https:\/\/mathority.org\/it\/esempi-di-regole-ed-esercizi-risolti-di-cramer\/","title":{"rendered":"Regola di cramer"},"content":{"rendered":"<p>In questa pagina vedrai cos&#8217;\u00e8 la regola di Cramer e, inoltre, troverai esempi ed esercizi con la risoluzione di sistemi di equazioni secondo la regola di Cramer.<\/p>\n<h2 class=\"wp-block-heading\"> Qual \u00e8 la regola di Cramer?<\/h2>\n<p> <strong>La regola di Cramer<\/strong> \u00e8 un metodo utilizzato per risolvere sistemi di equazioni mediante determinanti. Vediamo come si usa:<\/p>\n<p> Consideriamo un sistema di equazioni:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e0141f3451719f665ef28e4061489551_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} ax+by+cz= \\color{red}\\bm{j} \\\\[1.5ex] dx+ey+fz=\\color{red}\\bm{k} \\\\[1.5ex] gx+hy+iz = \\color{red}\\bm{l} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"171\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> La matrice A e la matrice estesa A&#8217; del sistema sono:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1d628a13ec7de4b3ba7a301c0a5d8ac6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} a &amp; b &amp; c  \\\\[1.1ex] d &amp; e &amp; f  \\\\[1.1ex] g &amp; h &amp; i  \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} a &amp; b &amp; c &amp;  \\color{red}\\bm{j}  \\\\[1.1ex] d &amp; e &amp; f &amp; \\color{red}\\bm{k} \\\\[1.1ex] g &amp; h &amp; i &amp; \\color{red}\\bm{l} \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"384\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong><span style=\"text-decoration: underline;\">La regola di Cramer<\/span><\/strong> afferma che la soluzione di un sistema di equazioni \u00e8: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/quelle-est-la-regle-de-cramer.webp\" alt=\"cos'\u00e8 la regola di Cramer, spiegazione della regola di Cramer\" class=\"wp-image-1062\" width=\"677\" height=\"385\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Si noti che i determinanti dei numeratori sono come il determinante della matrice A ma cambiando la colonna di ciascuna incognita nella colonna dei termini indipendenti.<\/p>\n<p> Pertanto, la regola di Cramer viene utilizzata per risolvere sistemi di equazioni lineari. Ma, come gi\u00e0 sai, ci sono molti modi per risolvere un sistema di equazioni, ad esempio <a href=\"https:\/\/mathority.org\/it\/metodo-jordan-gauss-con-esempi-ed-esercizi-risolti\/\">il metodo di Gauss Jordan<\/a> \u00e8 ben noto.<\/p>\n<p> Di seguito sono riportati esempi di risoluzione di sistemi di equazioni lineari con la regola di Cramer, o talvolta anche scritti come regola di Kramer.<\/p>\n<h2 class=\"wp-block-heading\"> Esempio 1: sistema compatibile determinato (SCD)<\/h2>\n<ul>\n<li> Risolvi il seguente sistema di 3 equazioni in 3 incognite utilizzando la regola di Cramer:<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6013b7e73c89c24fe388f1a5d018f32b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 2x+y+3z= 1 \\\\[1.5ex] 3x-2y-z=0 \\\\[1.5ex] x+3y+2z = 5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"135\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Per prima cosa realizziamo la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c710ed86223f47f39b5a25720b5ca19d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; 3 &amp; 1 \\\\[1.1ex] 3 &amp; -2 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 3 &amp; 2 &amp; 5 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Calcoliamo ora il rango delle due matrici, per vedere di che tipo di sistema si tratta. Per calcolare il rango di A, calcoliamo il determinante 3\u00d73 dell&#8217;intera matrice (usando la regola di Sarrus) e vediamo se d\u00e0 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ae4a3bb88d113494463df8e670c326c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{vmatrix} =-8-1+27+6+6-6 = 24 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"427\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Il determinante di A \u00e8 diverso da 0, quindi <strong>la matrice A ha rango 3.<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Quindi <strong>anche la matrice A&#8217; \u00e8 di rango 3<\/strong> , poich\u00e9 non pu\u00f2 essere di rango 4 e deve essere almeno dello stesso rango della matrice A.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> L&#8217;estensione della matrice A \u00e8 pari all&#8217;estensione della matrice A&#8217; e al numero di incognite del sistema (3), quindi, per il <strong>teorema di Rouch\u00e9-Frobenius<\/strong> , sappiamo che si tratta di un <strong>determinato sistema compatibile<\/strong> (SCD):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-557185e16670c72d23eec5a3ea13b487_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Una volta che sappiamo che il sistema \u00e8 una SCD, applichiamo <strong>la regola di Cramer<\/strong> per risolverlo. Per fare ci\u00f2 ricordiamo che la matrice A, il suo determinante e la matrice A&#8217; sono:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b2b3e5865c2264c360fb887d37a5f6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 2 &amp; 1 &amp; 3 &amp; \\color{red}\\bm{1} \\\\[1.1ex] 3 &amp; -2 &amp; -1 &amp; \\color{red}\\bm{0} \\\\[1.1ex] 1 &amp; 3 &amp; 2 &amp; \\color{red}\\bm{5} \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"431\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0a604d8f5a3927a47a264d28f7a007b2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 1 &amp; 3 \\\\[1.1ex] 3 &amp; -2 &amp; -1 \\\\[1.1ex] 1 &amp; 3 &amp; 2\\end{vmatrix} =24\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"187\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a1fa494ffb5e452d59c4d2dad40f925a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} \\color{red}\\bm{1} &amp; 1 &amp; 3 \\\\[1.1ex] \\color{red}\\bm{0} &amp; -2 &amp; -1 \\\\[1.1ex] \\color{red}\\bm{5} &amp; 3 &amp; 2 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{24}{24} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"238\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-08e3dabe2f33434eb96658491f67c0b4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 2 &amp; \\color{red}\\bm{1} &amp; 3 \\\\[1.1ex] 3 &amp;  \\color{red}\\bm{0} &amp; -1 \\\\[1.1ex] 1 &amp; \\color{red}\\bm{5} &amp; 2\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{48}{24} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"223\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Calcolare<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la terza colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96e76cb8867224755e9c19254678abd4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} 2 &amp; 1 &amp; \\color{red}\\bm{1} \\\\[1.1ex] 3 &amp; -2 &amp;  \\color{red}\\bm{0} \\\\[1.1ex] 1 &amp; 3 &amp;  \\color{red}\\bm{5}\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-24}{24} = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"259\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> La soluzione del sistema di equazioni \u00e8 quindi:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-be5a19fed42dcb59880c2d0eee8e51f4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = 1 \\qquad y=2 \\qquad z = -1}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"210\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> Esempio 2: sistema compatibile indeterminato (ICS)<\/h2>\n<ul>\n<li> Risolvi il seguente sistema di equazioni utilizzando la regola di Cramer:<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-781530aac4d8507fd6c7cbd77c3b4651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0 \\\\[1.5ex] x+5y+3z = 1 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"149\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Per prima cosa realizziamo la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a64800a78bf8e2e2f547be907e6863cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; 3 &amp; -1 \\\\[1.1ex] 1 &amp; 5 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; 2 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 5 &amp; 3 &amp; 1 \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"405\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Ora calcoliamo il range delle due matrici e quindi possiamo vedere di che tipo di sistema si tratta. Per calcolare il rango di A, calcoliamo il determinante dell&#8217;intera matrice (usando la regola di Sarrus) e controlliamo se \u00e8 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-581c58cbe0fdd9952e7e25b919ecc33b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; 3 &amp; -1 \\\\[1.1ex] 1 &amp; 5 &amp; 3\\end{vmatrix} = 27-2-40-12+15+12= 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"407\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Il determinante d\u00e0 0, quindi la matrice A non \u00e8 di rango 3. Ma ha un determinante 2\u00d72 diverso da 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a5d1acad8bc31240f80d8cfbf3605997_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; 2 \\\\[1.1ex] -2 &amp; 3 \\end{vmatrix} =9-(-4)=13\\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Quindi <strong>la matrice A ha rango 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Una volta conosciuta l&#8217;estensione della matrice A, calcoliamo quella della matrice A&#8217;. Il determinante delle prime 3 colonne d\u00e0 0, quindi proviamo gli altri possibili determinanti 3\u00d73 nella matrice A&#8217;:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-686e7ca635ecee685005f6013c2e64ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 4 &amp; 1 \\\\[1.1ex] 3 &amp; -1 &amp; 0 \\\\[1.1ex] 5 &amp; 3 &amp; 1 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; 4 &amp; 1 \\\\[1.1ex] -2 &amp; -1 &amp; 0 \\\\[1.1ex] 1 &amp; 3 &amp; 1 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; 2 &amp; 1 \\\\[1.1ex] -2 &amp; 3 &amp; 0 \\\\[1.1ex] 1 &amp; 5 &amp; 1 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"440\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Tutti i determinanti di ordine 3 danno 0. Ma, ovviamente, la matrice A&#8217; ha lo stesso determinante 2\u00d72 non 0 della matrice A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a5d1acad8bc31240f80d8cfbf3605997_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; 2 \\\\[1.1ex] -2 &amp; 3 \\end{vmatrix} =9-(-4)=13\\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"222\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Pertanto <strong>anche la matrice A&#8217; \u00e8 di rango 2<\/strong> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Quindi, poich\u00e9 il rango della matrice A \u00e8 uguale al rango della matrice A&#8217; ma questi due sono inferiori al numero di incognite del sistema (3), sappiamo dal <strong>teorema di Rouch\u00e9-Frobenius<\/strong> che \u00e8 Questo \u00e8 un <strong>sistema indeterminatamente compatibile<\/strong> (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-background\" style=\"background-color:#dff6ff\"> Quando vogliamo risolvere un sistema indeterminato compatibile (SCI), dobbiamo <strong>trasformare il sistema<\/strong> : prima eliminiamo un&#8217;equazione, poi convertiamo una variabile in \u03bb (solitamente la variabile z), e infine mettiamo insieme i termini con \u03bb con i termini indipendenti.<\/p>\n<p> Una volta trasformato il sistema applichiamo la regola di Cramer e otterremo la soluzione del sistema in funzione di \u03bb.<\/p>\n<p> In questo caso <strong>elimineremo l\u2019ultima equazione<\/strong> dal sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f0511fecc9c2af695b6b8eccae6b0661_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0 \\\\[1.5ex]\\cancel{x+5y+3z = 1} \\end{cases} \\longrightarrow \\quad \\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <strong>Ora convertiamo la variabile z in \u03bb:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2d6142d2be611954fd849a032a97245a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y+4z=1 \\\\[1.5ex] -2x+3y-z=0  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 3x+2y+4\\lambda=1 \\\\[1.5ex] -2x+3y-\\lambda=0\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> E mettiamo <strong>i termini con \u03bb con i termini indipendenti:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-00214205f2334f1c9bc10810c1c1df83_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x+2y=1-4\\lambda \\\\[1.5ex] -2x+3y=\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"adsb30\" style=\" margin:px; text-align:\"><\/div>\n<p> Pertanto la matrice A e la matrice A&#8217; del sistema restano:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9c4b47303973b823a1c5628f5448ca79_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; 2  \\\\[1.1ex] -2 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 3 &amp; 2 &amp; 1 -4\\lambda \\\\[1.1ex] -2 &amp; 3 &amp; \\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"363\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Infine, una volta trasformato il sistema, <strong>applichiamo la regola di Cramer<\/strong> . Risolviamo quindi il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d1b79f52dc82f5cfc311867273e78c06_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; 2  \\\\[1.1ex] -2 &amp; 3\\end{vmatrix} = 13\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"148\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0ff917eaea976c65bd18e0476078d3cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 1 -4\\lambda &amp; 2  \\\\[1.1ex] \\lambda &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3(1-4\\lambda) -2\\lambda}{13} = \\cfrac{\\bm{3-14\\lambda} }{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"349\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-155ca520739bbf7e040a6cdc632f7c27_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 3 &amp; 1 -4\\lambda  \\\\[1.1ex]-2&amp;  \\lambda  \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3\\lambda -\\bigl(-2(1-4\\lambda)\\bigr)}{13}= \\cfrac{3\\lambda -\\bigl(-2+8\\lambda\\bigr)}{13} = \\cfrac{\\bm{2-5\\lambda} }{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"529\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Mentre la soluzione del sistema di equazioni \u00e8 funzione di \u03bb, poich\u00e9 \u00e8 uno SCI e, quindi, ha infinite soluzioni:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c9866e045041eb2d8fe103db2309f229_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =} \\cfrac{\\bm{3-14\\lambda} }{\\bm{13}} \\qquad \\bm{y=}\\cfrac{\\bm{2-5\\lambda} }{\\bm{13}} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"283\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"> La regola di Cramer ha risolto i problemi <\/h2>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-118\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Esercizio 1<\/h3>\n<p> Applica la regola di Cramer per risolvere il seguente sistema di due equazioni in 2 incognite: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-pas-a-pas-de-la-regle-de-cramer-22.webp\" alt=\"esercizio risolto passo dopo passo con la regola 2x2 di Cramer\" class=\"wp-image-3999\" width=\"137\" height=\"83\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>vedi soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> La prima cosa da fare \u00e8 la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2a001db9cf56846150730fee7126dacd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{cc} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 2 &amp; 5 &amp; 8 \\\\[1.1ex] 1 &amp; 4 &amp; 7 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"294\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dobbiamo ora trovare il rango della matrice A. Per fare ci\u00f2 controlliamo se il determinante dell&#8217;intera matrice \u00e8 diverso da 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0c75c1c344c286016bea83237f1f418e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 2 &amp; 5 \\\\[1.1ex] 1 &amp; 4 \\end{vmatrix} = 8-5=3 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"216\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Poich\u00e9 la matrice ha un determinante 2\u00d72 diverso da 0, <strong>la matrice A ha rango 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta conosciuto il rango di A, calcoliamo il rango di A&#8217;. Questo sar\u00e0 almeno di rango 2, perch\u00e9 abbiamo appena visto che ha al suo interno un determinante di ordine 2 diverso da 0. Inoltre, non pu\u00f2 essere di rango 3, poich\u00e9 non possiamo fare a meno di un determinante 3\u00d73. Pertanto <strong>anche la matrice A&#8217; \u00e8 di rango 2:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto, applicando il <strong>teorema di Rouch\u00e9-Frobenius,<\/strong> sappiamo che questo \u00e8 un <strong>sistema determinato compatibile<\/strong> (SCD), perch\u00e9 l&#8217;intervallo di A \u00e8 uguale all&#8217;intervallo di A&#8217; e al numero di incognite.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bbd67b16bb6d52a0696e70a77833cd3b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 2 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 2 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta che sappiamo che il sistema \u00e8 una SCD, applichiamo <strong>la regola di Cramer<\/strong> per risolverlo.<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b0adeda8f2ce557661466996038b1148_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 8 &amp; 5 \\\\[1.1ex] 7 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-3}{3} = \\bm{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-59790a66cc31fac07be1d5a7bb556d9e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}2 &amp; 8 \\\\[1.1ex] 1 &amp; 7\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{6}{3} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"150\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> La soluzione del sistema di equazioni \u00e8 quindi: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-26fa7c9ed2d05ca07ff62a968ba7ab11_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x = -1 \\qquad y=2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"133\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Esercizio 2<\/h3>\n<p> Trova la soluzione del seguente sistema di tre equazioni in 3 incognite utilizzando la regola di Cramer: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exercice-resolu-du-systeme-de-regles-de-cramer-des-equations-3-3.webp\" alt=\"Esercizio risolto della regola di Cramer di un sistema di equazioni 3x3\" class=\"wp-image-4002\" width=\"181\" height=\"124\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>vedi soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Per prima cosa realizziamo la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eea75fbf6d86ebc3d0b9e236cd2160f5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 3 &amp; 2\\\\[1.1ex] -1 &amp; 5 &amp; -1\\\\[1.1ex] 3 &amp; -1 &amp; 4 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 3 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; -1 &amp; 4 \\\\[1.1ex] 3 &amp; -1 &amp; 4 &amp; 0 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"432\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Troviamo ora il rango della matrice A calcolando il determinante della matrice 3\u00d73 con la regola di Sarrus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-73f751f3b5c527c16b5de1b10bf07a4e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 3 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; -1\\\\[1.1ex] 3 &amp; -1 &amp; 4 \\end{vmatrix} = 20-9+2-30-1+12=-6 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"445\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> La matrice avente un determinante di ordine 3 diverso da 0, <strong>la matrice A \u00e8 di rango 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> di conseguenza anche la matrice A&#8217; \u00e8 di rango 3:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto, utilizzando il <strong>teorema di Rouch\u00e9-Frobenius,<\/strong> sappiamo che questo \u00e8 un <strong>sistema determinato compatibile<\/strong> (SCD), perch\u00e9 l&#8217;intervallo di A \u00e8 uguale all&#8217;intervallo di A&#8217; e al numero di incognite.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b495a48a75d7af1f23e38818bf4eca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta che sappiamo che il sistema \u00e8 una SCD, dobbiamo applicare <strong>la regola di Cramer<\/strong> per risolverlo.<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fc574297f609b68e4fb48466ec6c8077_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 2 &amp; 3 &amp; 2 \\\\[1.1ex] 4 &amp; 5 &amp; -1\\\\[1.1ex]0 &amp; -1 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-18}{-6} = \\bm{3}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"235\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2544601137d62e217ff1866f278203d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}1 &amp; 2 &amp; 2 \\\\[1.1ex] -1 &amp; 4 &amp; -1\\\\[1.1ex] 3 &amp; 0 &amp; 4\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-6}{-6} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"224\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcolare<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la terza colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-42d7d4adcfc48954185ca14b56b8e128_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} 1 &amp; 3 &amp; 2 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 3 &amp; -1 &amp; 0\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{12}{-6} = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"230\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> La soluzione del sistema di equazioni \u00e8 quindi: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-685195d3a299f30f6421bb387f7f00e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =3 \\qquad y=1 \\qquad z=-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"210\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Esercizio 3<\/h3>\n<p> Calcola la soluzione del seguente sistema di tre equazioni in 3 incognite utilizzando la regola di Cramer: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-de-regle-de-cramer.webp\" alt=\"esempio della regola di Cramer\" class=\"wp-image-4003\" width=\"183\" height=\"123\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>vedi soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Per prima cosa realizziamo la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-afd359275e5ebaaf3229504c47a5815f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2 &amp; 5\\\\[1.1ex] 2 &amp; 3 &amp; -1 \\\\[1.1ex] 3 &amp; 4 &amp; -7 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 1 &amp; 2 &amp; 5 &amp; 1 \\\\[1.1ex] 2 &amp; 3 &amp; -1 &amp; 5 \\\\[1.1ex] 3 &amp; 4 &amp; -7 &amp; 9 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"377\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcoliamo l&#8217;estensione della matrice A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-47ddf17a2b3eed5a680d685900a79b31_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 2 &amp; 5\\\\[1.1ex] 2 &amp; 3 &amp; -1 \\\\[1.1ex] 3 &amp; 4 &amp; -7 \\end{vmatrix} =-21-6+40-45+4+28=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"398\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fdd4380c7c76418bd3ec12c94359f886_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3  \\end{vmatrix} = 3-4 = -1 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta conosciuta l&#8217;estensione della matrice A, calcoliamo quella della matrice A&#8217;. Il determinante delle prime 3 colonne d\u00e0 0, quindi proviamo gli altri possibili determinanti 3\u00d73 nella matrice A&#8217;:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1addc62130e0462075b3bade26a7e35e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 2 &amp; 5 &amp; 1 \\\\[1.1ex]  3 &amp; -1 &amp; 5 \\\\[1.1ex] 4 &amp; -7 &amp; 9 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 5 &amp; 1 \\\\[1.1ex] 2 &amp; -1 &amp; 5 \\\\[1.1ex] 3 &amp; -7 &amp; 9\\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 1 &amp; 2 &amp; 1 \\\\[1.1ex] 2 &amp; 3 &amp; 5 \\\\[1.1ex] 3 &amp; 4 &amp; 9 \\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"412\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tutti i determinanti di ordine 3 danno 0. Tuttavia, la matrice A&#8217; ha lo stesso determinante 2\u00d72 non 0 della matrice A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7de377466bd5afd03f58f9b532324e75_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3 \\end{vmatrix} = 3-4 = -1 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"186\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto anche la matrice A&#8217; \u00e8 di rango 2:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Poich\u00e9 il rango della matrice A \u00e8 uguale al rango della matrice A&#8217; ma questi due sono inferiori al numero di incognite del sistema (3), sappiamo dal <strong>teorema di Rouch\u00e9-Frobenius<\/strong> che si tratta di un <strong>sistema compatibile indeterminato<\/strong> (ICS):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Essendo un sistema ICS, dobbiamo eliminare un&#8217;equazione. In questo caso <strong>elimineremo l\u2019ultima equazione<\/strong> dal sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3a1d067e155540f4345cf56e5c1567d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5 \\\\[1.5ex]\\cancel{3x+4y-7z = 9} \\end{cases} \\longrightarrow \\quad \\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"357\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Ora convertiamo la variabile z in \u03bb:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b5fa91777a722d3783b2f887aab44152_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y+5z=1 \\\\[1.5ex] 2x+3y-z=5  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} x+2y+5\\lambda=1 \\\\[1.5ex] 2x+3y-\\lambda=5\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"369\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E mettiamo i termini con \u03bb con i termini indipendenti:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-76ff21181be050b01c247981298986a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} x+2y=1-5\\lambda\\\\[1.5ex] 2x+3y=5+\\lambda \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"136\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tali che la matrice A e la matrice A&#8217; del sistema rimangono:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-230e5b28dd467127e63f4f9756cf90da_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 1 &amp; 2  \\\\[1.1ex] 2 &amp; 3 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 1 &amp; 2 &amp; 1 -5\\lambda \\\\[1.1ex] 2 &amp; 3 &amp;5+\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"335\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Infine, una volta trasformato il sistema, <strong>applichiamo la regola di Cramer<\/strong> . Risolviamo quindi il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f127efbd217e2bca8852ec792610732f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 1 &amp; 2 \\\\[1.1ex] 2 &amp; 3\\end{vmatrix} =-1\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"138\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-42652a14362b42e606841b6bb3e77cc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 1-5\\lambda &amp; 2 \\\\[1.1ex] 5+\\lambda &amp; 3 \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{3-15\\lambda -(10+2\\lambda)}{-1} = \\cfrac{-7-17\\lambda}{-1} = \\bm{7+17\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"491\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b95c5870f1762a2d82c9ebcccbca7408_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 1 &amp; 1-5\\lambda \\\\[1.1ex] 2 &amp; 5+\\lambda \\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{5+\\lambda -(2-10\\lambda)}{-1}= \\cfrac{3+11\\lambda}{-1} = \\bm{-3-11\\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"465\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mentre la soluzione del sistema di equazioni \u00e8 funzione di \u03bb, poich\u00e9 \u00e8 uno SCI e, quindi, ha infinite soluzioni: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5483357f081aca551b07fe7c8f9ebf5d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =7+17\\lambda} \\qquad \\bm{y=-3-11\\lambda} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"311\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-119\"><\/div>\n<\/div>\n<h3 class=\"wp-block-heading\"> Esercizio 4<\/h3>\n<p> Risolvi il seguente problema di un sistema di tre equazioni in 3 incognite applicando la regola di Cramer: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-61e1c3458f33b863db10750b9e51d09e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} -2x+5y+z=8 \\\\[1.5ex] 6x+2y+4z=4 \\\\[1.5ex] 3x-2y+z = -2 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"149\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>vedi soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Per prima cosa costruiamo la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-743a40010cb4a610e8a3fc6ae5d313b4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc}-2 &amp; 5 &amp; 1 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1\\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} -2 &amp; 5 &amp; 1 &amp; 8 \\\\[1.1ex] 6 &amp; 2 &amp; 4 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1 &amp; -2 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"419\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcoliamo ora il rango della matrice A calcolando il determinante della matrice 3&#215;3 utilizzando la regola di Sarrus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-713c634fbc3e1b1cb228e3891c9bff1c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} -2 &amp; 5 &amp; 1 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1 \\end{vmatrix} = -4+60-12-6-16-30=-8 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"453\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> La matrice avente un determinante di ordine 3 diverso da 0, <strong>la matrice A \u00e8 di rango 3:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-842ae3b68df41813d9e409968f3ae946_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> di conseguenza, anche la matrice A&#8217; \u00e8 di rango 3, poich\u00e9 deve essere almeno dello stesso rango della matrice A e non pu\u00f2 essere di rango 4 perch\u00e9 \u00e8 una matrice di dimensione 3\u00d74.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-150bbc9c8e363db471c2d5bc4f33e1fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto, utilizzando il <strong>teorema di Rouch\u00e9-Frobenius,<\/strong> deduciamo che si tratta di un <strong>determinato sistema compatibile<\/strong> (SCD), perch\u00e9 l&#8217;intervallo di A \u00e8 uguale all&#8217;intervallo di A&#8217; e al numero di incognite.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-31b495a48a75d7af1f23e38818bf4eca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 3 \\\\[1.3ex] \\color{black}rg(A')=3 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3 \\end{array}} \\\\ \\\\ \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = n = 3 \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCD}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"436\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta che sappiamo che il sistema \u00e8 una SCD, dobbiamo applicare <strong>la regola di Cramer<\/strong> per risolverlo.<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8a290479c69ff806f19dcf29f96e1228_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 8 &amp; 5 &amp; 1 \\\\[1.1ex] 4 &amp; 2 &amp; 4 \\\\[1.1ex] -2 &amp; -2 &amp; 1\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{16}{-8} = \\bm{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, cambiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8bba0765fbcbcebf0585520af25b4a30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix}-2 &amp; 8 &amp; 1 \\\\[1.1ex] 6 &amp; 4 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; 1\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{0}{-6} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"217\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcolare<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23aa090e6102a41de5ad5515112e4d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  z\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la terza colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5bc157a8c4dfe8ee4651affac68ef878_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{z} = \\cfrac{\\begin{vmatrix} -2 &amp; 5 &amp; 8 \\\\[1.1ex] 6 &amp; 2 &amp; 4 \\\\[1.1ex] 3 &amp; -2 &amp; -2\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-32}{-8} = \\bm{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"113\" width=\"247\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> La soluzione del sistema di equazioni lineari \u00e8 quindi: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6c004c5c466235d2d1a784707145d952_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x =-2 \\qquad y=0 \\qquad z=4}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"211\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Esercizio 5<\/h3>\n<p> Risolvi il seguente sistema di equazioni lineari utilizzando la regola di Cramer: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/exemple-comment-resoudre-un-systeme-dequations-avec-la-regle-de-cramer.webp\" alt=\"Esempio di risoluzione di un sistema di equazioni con la regola di Cramer\" class=\"wp-image-4008\" width=\"215\" height=\"127\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E3F2FD boto_ver_solucion\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E3F2FD\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>vedi soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Per prima cosa realizziamo la matrice A e la matrice estesa A&#8217; del sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f5153b5951b768cc3cafa2bb2567ba92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; -2 &amp; -3 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 5 &amp; 1 &amp; -2 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{ccc|c} 3 &amp; -2 &amp; -3 &amp; 4 \\\\[1.1ex] -1 &amp; 5 &amp; 4 &amp; -10 \\\\[1.1ex] 5 &amp; 1 &amp; -2 &amp; -2 \\end{array}\\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"85\" width=\"455\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcoliamo l&#8217;estensione della matrice A: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f3778c9499e2a44ea3834dfed1523163_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix} 3 &amp; -2 &amp; -3 \\\\[1.1ex] -1 &amp; 5 &amp; 4 \\\\[1.1ex] 5 &amp; 1 &amp; -2 \\end{vmatrix} =-30-40+3+75-12+4=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"426\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-03d70742b14ced92f33963df0c86e92f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -2 \\\\[1.1ex] -1 &amp; 5  \\end{vmatrix} = 15- (2)= 13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-eded270b78ab3d95ce827e3ea428efb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta conosciuta l&#8217;estensione della matrice A, calcoliamo quella della matrice A&#8217;. Il determinante delle prime 3 colonne d\u00e0 0, quindi proviamo gli altri possibili determinanti 3\u00d73 nella matrice A&#8217;:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5bed93d532ae4ccd4649a73662f55f0f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} -2 &amp; -3 &amp; 4 \\\\[1.1ex] 5 &amp; 4 &amp; -10 \\\\[1.1ex]  1 &amp; -2 &amp; -2 \\end{vmatrix} = 0 \\qquad \\begin{vmatrix}3 &amp; -3 &amp; 4 \\\\[1.1ex] -1 &amp; 4 &amp; -10 \\\\[1.1ex] 5 &amp; -2 &amp; -2\\end{vmatrix} = 0 \\qquad \\begin{vmatrix} 3 &amp; -2 &amp; 4 \\\\[1.1ex] -1 &amp; 5 &amp; -10 \\\\[1.1ex] 5 &amp; 1 &amp;-2\\end{vmatrix} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"535\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tutti i determinanti di ordine 3 danno 0. Ma, ovviamente, la matrice A&#8217; ha lo stesso determinante di ordine 2 diverso da 0 della matrice A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-858d95d7d252b16706b66c0e6aba09c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix} 3 &amp; -2 \\\\[1.1ex] -1 &amp; 5 \\end{vmatrix} = 13 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"145\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto anche la matrice A&#8217; \u00e8 di rango 2:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80398cfd2fff647f81c0d4160f3b2f7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  rg(A')=2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"81\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Il rango della matrice A \u00e8 uguale al rango della matrice A&#8217; ma questi due sono inferiori al numero di incognite del sistema (3), quindi dal <strong>teorema di Rouch\u00e9-Frobenius<\/strong> sappiamo che \u00e8 un <strong>Sistema Indeterminato Compatibile<\/strong> (SCI) :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-96868a2569ea0ab5ca99d8dc606d3dc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{array}{c} \\begin{array}{c} \\color{black}rg(A) = 2 \\\\[1.3ex] \\color{black}rg(A')=2 \\\\[1.3ex] \\color{black}\\text{N\\'umero de inc\\'ognitas} = 3    \\end{array}} \\\\ \\\\  \\color{blue} \\boxed{ \\color{black}\\phantom{^9_9} rg(A) = rg(A') = 2 \\ < \\ n =3  \\color{blue} \\ \\bm{\\longrightarrow} \\ \\color{black} \\bm{SCI}\\phantom{^9_9}} \\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"138\" width=\"475\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Essendo un sistema ICS, dobbiamo eliminare un&#8217;equazione. In questo caso <strong>elimineremo l\u2019ultima equazione<\/strong> dal sistema:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4e10bd826663dff41c4272610cbc07b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10 \\\\[1.5ex]\\cancel{5x+y-2z = -2} \\end{cases} \\longrightarrow \\quad \\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"97\" width=\"423\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Ora convertiamo la variabile z in \u03bb:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2502be450040b38761c08e5d6beaf379_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y-3z=4 \\\\[1.5ex] -x+5y+4z=-10  \\end{cases} \\xrightarrow{z \\ = \\ \\lambda}\\quad \\begin{cases} 3x-2y-3\\lambda=4 \\\\[1.5ex] -x+5y+4\\lambda=-10\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"444\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E mettiamo i termini con \u03bb con i termini indipendenti:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-80a43d98e6be30965d554e8a89aa5d89_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases} 3x-2y=4+3\\lambda \\\\[1.5ex] -x+5y=-10-4\\lambda\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"172\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tali che la matrice A e la matrice A&#8217; del sistema rimangono:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3451ce571163983cf41794d4998283d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  A= \\left( \\begin{array}{ccc} 3 &amp; -2  \\\\[1.1ex] -1 &amp; 5 \\end{array} \\right) \\qquad A'= \\left( \\begin{array}{cc|c} 3 &amp; -2 &amp; 4+3\\lambda \\\\[1.1ex] 1 &amp; 5 &amp;-10-4\\lambda \\end{array} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"399\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Infine, una volta trasformato il sistema, <strong>applichiamo la regola di Cramer<\/strong> . Risolviamo quindi il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0e7a7d6208ea5e762f5c74a44e6838cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\begin{vmatrix}A \\end{vmatrix}= \\begin{vmatrix}3&amp; -2 \\\\[1.1ex] -1 &amp; 5\\end{vmatrix} =13\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"162\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-757d0eed520b26d08cc3b8b397d0f980_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> Con la regola di Cramer, sostituiamo la prima colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8c30fcc0526c2d4112eb4f60a3d8847f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x} = \\cfrac{\\begin{vmatrix} 4+3\\lambda &amp; -2 \\\\[1.1ex]-10-4\\lambda &amp; 5\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{20+15\\lambda -(20+8\\lambda)}{13} = \\cfrac{\\bm{7\\lambda}}{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"394\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Per calcolare l&#8217;incognita<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5fb4fb8b1addff607711094fd1ed326e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> Con la regola di Cramer, cambiamo la seconda colonna del determinante di A con la colonna dei termini indipendenti e la dividiamo per il determinante di A:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fdb22a54274e019c811c9051502c474a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{y} = \\cfrac{\\begin{vmatrix} 3 &amp; 4+3\\lambda \\\\[1.1ex] -1 &amp; -10-4\\lambda\\end{vmatrix}}{\\begin{vmatrix} A \\end{vmatrix}} = \\cfrac{-30-12\\lambda -(-4-3\\lambda)}{13}= \\cfrac{\\bm{-26-9\\lambda}}{\\bm{13}}\" title=\"Rendered by QuickLaTeX.com\" height=\"81\" width=\"473\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto, la soluzione del sistema di equazioni \u00e8 una funzione di \u03bb, poich\u00e9 \u00e8 uno SCI e, quindi, il sistema ha infinite soluzioni:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d0e525b9aca6bd683491ab7950f039e3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  \\bm{x=} \\cfrac{\\bm{7\\lambda}}{\\bm{13}} \\qquad \\bm{y=} \\cfrac{\\bm{-26-9\\lambda}}{\\bm{13}} \\qquad \\bm{z = \\lambda}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"266\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>In questa pagina vedrai cos&#8217;\u00e8 la regola di Cramer e, inoltre, troverai esempi ed esercizi con la risoluzione di sistemi di equazioni secondo la regola di Cramer. Qual \u00e8 la regola di Cramer? La regola di Cramer \u00e8 un metodo utilizzato per risolvere sistemi di equazioni mediante determinanti. Vediamo come si usa: Consideriamo un sistema &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/it\/esempi-di-regole-ed-esercizi-risolti-di-cramer\/\"> <span class=\"screen-reader-text\">Regola di cramer<\/span> Leggi altro &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[19],"tags":[],"class_list":["post-307","post","type-post","status-publish","format-standard","hentry","category-sistemi-educativi"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Regola di Cramer: Mathority<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathority.org\/it\/esempi-di-regole-ed-esercizi-risolti-di-cramer\/\" \/>\n<meta property=\"og:locale\" content=\"it_IT\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Regola di Cramer: Mathority\" \/>\n<meta property=\"og:description\" content=\"In questa pagina vedrai cos&#8217;\u00e8 la regola di Cramer e, inoltre, troverai esempi ed esercizi con la risoluzione di sistemi di equazioni secondo la regola di Cramer. Qual \u00e8 la regola di Cramer? La regola di Cramer \u00e8 un metodo utilizzato per risolvere sistemi di equazioni mediante determinanti. 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