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Expert-verified Found in: Page 307 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Use Cramer's rule to solve the systems in Exercises 22 through 24.23. $|\begin{array}{ccc}5{x}_{1}& -3{x}_{2}& =1\\ -6{x}_{1}& +7{x}_{2}& =0\end{array}|$

Therefore,

$x=\left[\begin{array}{c}\frac{7}{17}\\ \frac{6}{17}\end{array}\right]$

See the step by step solution

## Step 1: Definition.

In matrices, Cramer's rule expresses the solution in terms of the determinants of the coefficient matrix (i.e., for a square matrix) and of matrices obtained from it by replacing one column by the column vector of right-hand-sides of the equations.

## Step 2: To solve.

Using Cramer's rule, we solve:

$\mathrm{det}\mathrm{A}=\left|\begin{array}{cc}5& -3\\ -6& 7\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{det}\mathrm{A}=17\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{1}=\frac{\left|\begin{array}{cc}1& -3\\ 0& 7\end{array}\right|}{17}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{1}=\frac{7}{17}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}=\frac{\left|\begin{array}{cc}5& 1\\ -6& 0\end{array}\right|}{17}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}=\frac{6}{17}$

Therefore,

$\mathrm{x}=\left[\begin{array}{c}\frac{7}{17}\\ \frac{6}{17}\end{array}\right]$ ### Want to see more solutions like these? 