We are given the matrix:A=[12351\u22123\u2212134]A = \\begin{bmatrix} 1 & 2 & 3 \\\\ 5 & 1 & -3 \\\\ -1 & 3 & 4 \\end{bmatrix}A=\u200b15\u22121\u200b213\u200b3\u221234\u200b\u200b<\/p>\n\n\n\n
and the system of equations:x1+2×2+3×3=25×1+x2\u22123×3=53\u2212x1+3×2+4×3=1\\begin{aligned} x_1 + 2x_2 + 3x_3 &= 2 \\\\ 5x_1 + x_2 – 3x_3 &= 53 \\\\ – x_1 + 3x_2 + 4x_3 &= 1 \\end{aligned}x1\u200b+2×2\u200b+3×3\u200b5×1\u200b+x2\u200b\u22123×3\u200b\u2212x1\u200b+3×2\u200b+4×3\u200b\u200b=2=53=1\u200b<\/p>\n\n\n\n
\n\n\n\n
Step 1: Find the determinant of matrix AAA<\/h3>\n\n\n\n
We use cofactor expansion along the first row:det(A)=1\u22c5\u22231\u2212334\u2223\u22122\u22c5\u22235\u22123\u221214\u2223+3\u22c5\u222351\u221213\u2223\\text{det}(A) = 1 \\cdot \\begin{vmatrix} 1 & -3 \\\\ 3 & 4 \\end{vmatrix} – 2 \\cdot \\begin{vmatrix} 5 & -3 \\\\ -1 & 4 \\end{vmatrix} + 3 \\cdot \\begin{vmatrix} 5 & 1 \\\\ -1 & 3 \\end{vmatrix}det(A)=1\u22c5\u200b13\u200b\u221234\u200b\u200b\u22122\u22c5\u200b5\u22121\u200b\u221234\u200b\u200b+3\u22c5\u200b5\u22121\u200b13\u200b\u200b<\/p>\n\n\n\n
Compute the 2\u00d72 determinants:<\/p>\n\n\n\n
- \n
- \u22231\u2212334\u2223=(1)(4)\u2212(\u22123)(3)=4+9=13\\begin{vmatrix} 1 & -3 \\\\ 3 & 4 \\end{vmatrix} = (1)(4) – (-3)(3) = 4 + 9 = 13\u200b13\u200b\u221234\u200b\u200b=(1)(4)\u2212(\u22123)(3)=4+9=13<\/li>\n\n\n\n
- \u22235\u22123\u221214\u2223=(5)(4)\u2212(\u22123)(\u22121)=20\u22123=17\\begin{vmatrix} 5 & -3 \\\\ -1 & 4 \\end{vmatrix} = (5)(4) – (-3)(-1) = 20 – 3 = 17\u200b5\u22121\u200b\u221234\u200b\u200b=(5)(4)\u2212(\u22123)(\u22121)=20\u22123=17<\/li>\n\n\n\n
- \u222351\u221213\u2223=(5)(3)\u2212(1)(\u22121)=15+1=16\\begin{vmatrix} 5 & 1 \\\\ -1 & 3 \\end{vmatrix} = (5)(3) – (1)(-1) = 15 + 1 = 16\u200b5\u22121\u200b13\u200b\u200b=(5)(3)\u2212(1)(\u22121)=15+1=16<\/li>\n<\/ul>\n\n\n\n
So,det(A)=1(13)\u22122(17)+3(16)=13\u221234+48=27\\text{det}(A) = 1(13) – 2(17) + 3(16) = 13 – 34 + 48 = 27det(A)=1(13)\u22122(17)+3(16)=13\u221234+48=27<\/p>\n\n\n\n
\n\n\n\nStep 2: Use Cramer’s Rule<\/h3>\n\n\n\n
Let:b\u20d7=[2531]\\vec{b} = \\begin{bmatrix} 2 \\\\ 53 \\\\ 1 \\end{bmatrix}b=\u200b2531\u200b\u200b<\/p>\n\n\n\n
To find each variable using Cramer\u2019s Rule:xi=det\u2061(Ai)det\u2061(A)x_i = \\frac{\\det(A_i)}{\\det(A)}xi\u200b=det(A)det(Ai\u200b)\u200b<\/p>\n\n\n\n
Where AiA_iAi\u200b is the matrix formed by replacing the i-th column of AAA with vector bbb.<\/p>\n\n\n\n
\n\n\n\nFind det\u2061(A1)\\det(A_1)det(A1\u200b)<\/h4>\n\n\n\n
Replace the first column with bbb:A1=[223531\u22123134]A_1 = \\begin{bmatrix} 2 & 2 & 3 \\\\ 53 & 1 & -3 \\\\ 1 & 3 & 4 \\end{bmatrix}A1\u200b=\u200b2531\u200b213\u200b3\u221234\u200b\u200bdet\u2061(A1)=2\u22c5\u22231\u2212334\u2223\u22122\u22c5\u222353\u2212314\u2223+3\u22c5\u222353113\u2223\\det(A_1) = 2 \\cdot \\begin{vmatrix} 1 & -3 \\\\ 3 & 4 \\end{vmatrix} – 2 \\cdot \\begin{vmatrix} 53 & -3 \\\\ 1 & 4 \\end{vmatrix} + 3 \\cdot \\begin{vmatrix} 53 & 1 \\\\ 1 & 3 \\end{vmatrix}det(A1\u200b)=2\u22c5\u200b13\u200b\u221234\u200b\u200b\u22122\u22c5\u200b531\u200b\u221234\u200b\u200b+3\u22c5\u200b531\u200b13\u200b\u200b<\/p>\n\n\n\n
- \n
- \u22231\u2212334\u2223=1(4)\u2212(\u22123)(3)=4+9=13\\begin{vmatrix} 1 & -3 \\\\ 3 & 4 \\end{vmatrix} = 1(4) – (-3)(3) = 4 + 9 = 13\u200b13\u200b\u221234\u200b\u200b=1(4)\u2212(\u22123)(3)=4+9=13<\/li>\n\n\n\n
- \u222353\u2212314\u2223=53(4)\u2212(\u22123)(1)=212+3=215\\begin{vmatrix} 53 & -3 \\\\ 1 & 4 \\end{vmatrix} = 53(4) – (-3)(1) = 212 + 3 = 215\u200b531\u200b\u221234\u200b\u200b=53(4)\u2212(\u22123)(1)=212+3=215<\/li>\n\n\n\n
- \u222353113\u2223=53(3)\u22121(1)=159\u22121=158\\begin{vmatrix} 53 & 1 \\\\ 1 & 3 \\end{vmatrix} = 53(3) – 1(1) = 159 – 1 = 158\u200b531\u200b13\u200b\u200b=53(3)\u22121(1)=159\u22121=158<\/li>\n<\/ul>\n\n\n\n
Then,det\u2061(A1)=2(13)\u22122(215)+3(158)=26\u2212430+474=70\\det(A_1) = 2(13) – 2(215) + 3(158) = 26 – 430 + 474 = 70det(A1\u200b)=2(13)\u22122(215)+3(158)=26\u2212430+474=70<\/p>\n\n\n\n
So,x1=7027x_1 = \\frac{70}{27}x1\u200b=2770\u200b<\/p>\n\n\n\n
\n\n\n\nFind det\u2061(A2)\\det(A_2)det(A2\u200b)<\/h4>\n\n\n\n
Replace second column with bbb:A2=[123553\u22123\u2212114]A_2 = \\begin{bmatrix} 1 & 2 & 3 \\\\ 5 & 53 & -3 \\\\ -1 & 1 & 4 \\end{bmatrix}A2\u200b=\u200b15\u22121\u200b2531\u200b3\u221234\u200b\u200bdet\u2061(A2)=1\u22c5\u222353\u2212314\u2223\u22122\u22c5\u22235\u22123\u221214\u2223+3\u22c5\u2223553\u221211\u2223\\det(A_2) = 1 \\cdot \\begin{vmatrix} 53 & -3 \\\\ 1 & 4 \\end{vmatrix} – 2 \\cdot \\begin{vmatrix} 5 & -3 \\\\ -1 & 4 \\end{vmatrix} + 3 \\cdot \\begin{vmatrix} 5 & 53 \\\\ -1 & 1 \\end{vmatrix}det(A2\u200b)=1\u22c5\u200b531\u200b\u221234\u200b\u200b\u22122\u22c5\u200b5\u22121\u200b\u221234\u200b\u200b+3\u22c5\u200b5\u22121\u200b531\u200b\u200b<\/p>\n\n\n\n
- \n
- \u222353\u2212314\u2223=53(4)\u2212(\u22123)(1)=212+3=215\\begin{vmatrix} 53 & -3 \\\\ 1 & 4 \\end{vmatrix} = 53(4) – (-3)(1) = 212 + 3 = 215\u200b531\u200b\u221234\u200b\u200b=53(4)\u2212(\u22123)(1)=212+3=215<\/li>\n\n\n\n
- \u22235\u22123\u221214\u2223=5(4)\u2212(\u22123)(\u22121)=20\u22123=17\\begin{vmatrix} 5 & -3 \\\\ -1 & 4 \\end{vmatrix} = 5(4) – (-3)(-1) = 20 – 3 = 17\u200b5\u22121\u200b\u221234\u200b\u200b=5(4)\u2212(\u22123)(\u22121)=20\u22123=17<\/li>\n\n\n\n
- \u2223553\u221211\u2223=5(1)\u221253(\u22121)=5+53=58\\begin{vmatrix} 5 & 53 \\\\ -1 & 1 \\end{vmatrix} = 5(1) – 53(-1) = 5 + 53 = 58\u200b5\u22121\u200b531\u200b\u200b=5(1)\u221253(\u22121)=5+53=58<\/li>\n<\/ul>\n\n\n\n
So,det\u2061(A2)=1(215)\u22122(17)+3(58)=215\u221234+174=355\\det(A_2) = 1(215) – 2(17) + 3(58) = 215 – 34 + 174 = 355det(A2\u200b)=1(215)\u22122(17)+3(58)=215\u221234+174=355×2=35527x_2 = \\frac{355}{27}x2\u200b=27355\u200b<\/p>\n\n\n\n
\n\n\n\nFind det\u2061(A3)\\det(A_3)det(A3\u200b)<\/h4>\n\n\n\n
Replace third column with bbb:A3=[1225153\u2212131]A_3 = \\begin{bmatrix} 1 & 2 & 2 \\\\ 5 & 1 & 53 \\\\ -1 & 3 & 1 \\end{bmatrix}A3\u200b=\u200b15\u22121\u200b213\u200b2531\u200b\u200bdet\u2061(A3)=1\u22c5\u222315331\u2223\u22122\u22c5\u2223553\u221211\u2223+2\u22c5\u222351\u221213\u2223\\det(A_3) = 1 \\cdot \\begin{vmatrix} 1 & 53 \\\\ 3 & 1 \\end{vmatrix} – 2 \\cdot \\begin{vmatrix} 5 & 53 \\\\ -1 & 1 \\end{vmatrix} + 2 \\cdot \\begin{vmatrix} 5 & 1 \\\\ -1 & 3 \\end{vmatrix}det(A3\u200b)=1\u22c5\u200b13\u200b531\u200b\u200b\u22122\u22c5\u200b5\u22121\u200b531\u200b\u200b+2\u22c5\u200b5\u22121\u200b13\u200b\u200b<\/p>\n\n\n\n
- \n
- \u222315331\u2223=1(1)\u221253(3)=1\u2212159=\u2212158\\begin{vmatrix} 1 & 53 \\\\ 3 & 1 \\end{vmatrix} = 1(1) – 53(3) = 1 – 159 = -158\u200b13\u200b531\u200b\u200b=1(1)\u221253(3)=1\u2212159=\u2212158<\/li>\n\n\n\n
- \u2223553\u221211\u2223=5(1)\u221253(\u22121)=5+53=58\\begin{vmatrix} 5 & 53 \\\\ -1 & 1 \\end{vmatrix} = 5(1) – 53(-1) = 5 + 53 = 58\u200b5\u22121\u200b531\u200b\u200b=5(1)\u221253(\u22121)=5+53=58<\/li>\n\n\n\n
- \u222351\u221213\u2223=5(3)\u22121(\u22121)=15+1=16\\begin{vmatrix} 5 & 1 \\\\ -1 & 3 \\end{vmatrix} = 5(3) – 1(-1) = 15 + 1 = 16\u200b5\u22121\u200b13\u200b\u200b=5(3)\u22121(\u22121)=15+1=16<\/li>\n<\/ul>\n\n\n\n
So,det\u2061(A3)=1(\u2212158)\u22122(58)+2(16)=\u2212158\u2212116+32=\u2212242\\det(A_3) = 1(-158) – 2(58) + 2(16) = -158 – 116 + 32 = -242det(A3\u200b)=1(\u2212158)\u22122(58)+2(16)=\u2212158\u2212116+32=\u2212242×3=\u221224227x_3 = \\frac{-242}{27}x3\u200b=27\u2212242\u200b<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
x1=7027,x2=35527,x3=\u221224227\\boxed{ x_1 = \\frac{70}{27}, \\quad x_2 = \\frac{355}{27}, \\quad x_3 = \\frac{-242}{27} }x1\u200b=2770\u200b,x2\u200b=27355\u200b,x3\u200b=27\u2212242\u200b\u200b<\/p>\n\n\n\n
\n\n\n\nExplanation<\/h3>\n\n\n\n
Cramer\u2019s Rule is a method used to solve systems of linear equations using determinants. It is applicable only when the coefficient matrix has a nonzero determinant. In this problem, we are given a system of three equations with three variables. The coefficient matrix, denoted as AAA, is formed by extracting the coefficients of the variables from each equation. We first calculate the determinant of matrix AAA using cofactor expansion, which involves breaking the matrix into smaller 2×2 matrices and applying the formula for the determinant of a 3×3 matrix.<\/p>\n\n\n\n
After finding the determinant of the original matrix to be 27, we proceed to solve for each variable x1,x2,x_1, x_2,x1\u200b,x2\u200b, and x3x_3x3\u200b using Cramer\u2019s Rule. This involves constructing three new matrices A1,A2,A_1, A_2,A1\u200b,A2\u200b, and A3A_3A3\u200b by replacing one column of the original matrix with the constants from the right-hand side of the equations. Each of these modified matrices is then evaluated for its determinant.<\/p>\n\n\n\n
The values of the variables are found by dividing each new determinant by the determinant of the original matrix. This process highlights the power of determinants in solving linear systems, especially when traditional substitution or elimination methods are cumbersome. Though this method is not efficient for large systems due to its computational complexity, it provides a direct and exact solution for small systems, as shown here.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"1_ Find the determinant of the following matrix ~1 A = 5 ~1 2 ~1 3 3 -3 4 2 Use Cramer’s Rule to solve the following matrix ~X1 + 2X2 + 3×3 =2 5×1 X2 ~ 3×3 = 53 ~X1 + 3×2 + 4X3 =1 The Correct Answer and Explanation is: We are given […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","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":"","ast-disable-related-posts":"","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":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-238311","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238311","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=238311"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238311\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238311"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238311"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238311"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}