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: 1

Answer

The solutions to the system are: $x_{1}$ = $\frac{5}{6}$ $x_{2}$ = $\frac{-1}{6}$

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 = $\begin{bmatrix} 5 & 7 \\ 2 & 4 \end{bmatrix}$ We then replace the first column with the column of solutions. $A_{1}$(b) = $\begin{bmatrix} 3 & 7 \\ 1 & 4 \end{bmatrix}$ We then replace the second column with the column of solutions. $A_{2}$(b) = $\begin{bmatrix} 5 & 3 \\ 2 & 1 \end{bmatrix}$ 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)(4) - (7)(2) = 20 - 14 = 6 det($A_{1}$(b)) = (3)(4) - (7)(1) = 12 - 7 = 5 det($A_{2}$(b)) = (5)(1) - (3)(2) = 5 - 6 = -1 We then use the formula: $x_{1}$ = $\frac{det(A_{1}(b))}{det(A)}$ = $\frac{5}{6}$ $x_{2}$ = $\frac{det(A_{2}(b))}{det(A)}$ = $\frac{-1}{6}$ The solutions to the system are: $x_{1}$ = $\frac{5}{6}$ $x_{2}$ = $\frac{-1}{6}$
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