Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 3 - Determinants - 3.3 Exercises - Page 186: 4

Answer

$x_{1}$ = 1 $x_{2}$ = 7

Work Step by Step

Cramer's Rule is used to find the solutions to a system of equations. We first have to turn the system of equations into matrices. A = $\begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}$ We then replace the first column with the column of solutions. $A_{1}$(b) = $\begin{bmatrix} 9 & 2 \\ -4 & -1 \end{bmatrix}$ We then replace the second column with the column of solutions. $A_{2}$(b) = $\begin{bmatrix} -5 & 9 \\ 3 & -4 \end{bmatrix}$ We then need to find the determinant of each matrix. Note: The determinant of a 2x2 matrix is ad-bc where $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ det(A) = (-5)(-1) - (2)(3) = 5 - 6 = -1 det($A_{1}$(b)) = (9)(-1) - (-4)(2) = -9 + 8 = -1 det($A_{2}$(b)) = (-5)(-4) - (9)(3) = 20 - 27 = -7 We then use the formula: $x_{1}$ = $\frac{det(A_{1}(b))}{det(A)}$ = $\frac{-1}{-1}$ = 1 $x_{2}$ = $\frac{det(A_{2}(b))}{det(A)}$ = $\frac{-7}{-1}$ = 7 The solutions to the system are: $x_{1}$ = 1 $x_{2}$ = 7
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