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: 6

Answer

$x_{1}$ = $\frac{2}{5}$ $x_{2}$ = $\frac{4}{5}$ $x_{3}$ = $\frac{6}{5}$

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 = $\begin{bmatrix} 1 & 3 & 1 \\ -1 & 0 & 2\\ 3 & 1 & 0 \end{bmatrix}$ We then replace the first column with the column of solutions. $A_{1}$(b) = $\begin{bmatrix} 4 & 3 & 1 \\ 2 & 0 & 2\\ 2 & 1 & 0 \end{bmatrix}$ We then replace the second column with the column of solutions. $A_{2}$(b) = $\begin{bmatrix} 1 & 4 & 1 \\ -1 & 2 & 2\\ 3 & 2 & 0 \end{bmatrix}$ We then replace the third column with the column of solutions. $A_{3}$(b) = $\begin{bmatrix} 1 & 3 & 4 \\ -1 & 0 & 2\\ 3 & 1 & 2 \end{bmatrix}$ We then need to find the determinant of each matrix. det(A) = 15 det($A_{1}$(b)) = 6 det($A_{2}$(b)) = 12 det($A_{3}$(b)) = 18 We then use the formula: $x_{1}$ = $\frac{det(A_{1}(b))}{det(A)}$ = $\frac{6}{15}$ = $\frac{2}{5}$ $x_{2}$ = $\frac{det(A_{2}(b))}{det(A)}$ = $\frac{12}{15}$ = $\frac{4}{5}$ $x_{3}$ = $\frac{det(A_{3}(b))}{det(A)}$ = $\frac{18}{15}$ = $\frac{6}{5}$ The solutions to the system are: $x_{1}$ = $\frac{2}{5}$ $x_{2}$ = $\frac{4}{5}$ $x_{3}$ = $\frac{6}{5}$
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