Short Answer

Use Cramer's rule to solve the systems in Exercises 22 through 24.
23. $| \begin{matrix} 5 x_{1} & - 3 x_{2} & = 1 \\ - 6 x_{1} & + 7 x_{2} & = 0 \end{matrix} |$

Therefore,

$x = [\begin{matrix} \frac{7}{17} \\ \frac{6}{17} \end{matrix}]$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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:

$\det A = |\begin{matrix} 5 & - 3 \\ - 6 & 7 \end{matrix}| \det A = 17 x_{1} = \frac{|\begin{matrix} 1 & - 3 \\ 0 & 7 \end{matrix}|}{17} x_{1} = \frac{7}{17} x_{2} = \frac{|\begin{matrix} 5 & 1 \\ - 6 & 0 \end{matrix}|}{17} x_{2} = \frac{6}{17}$

Therefore,

$x = [\begin{matrix} \frac{7}{17} \\ \frac{6}{17} \end{matrix}]$

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Linear Algebra With Applications

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Short Answer

Use Cramer's rule to solve the systems in Exercises 22 through 24.
23. $| \begin{matrix} 5 x_{1} & - 3 x_{2} & = 1 \\ - 6 x_{1} & + 7 x_{2} & = 0 \end{matrix} |$

Step by Step Solution

Step 1: Definition.

Step 2: To solve.

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Linear Algebra With Applications

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Use Cramer's rule to solve the systems in Exercises 22 through 24.23. |5x1-3x2=1-6x1+7x2=0|

Step by Step Solution

Step 1: Definition.

Step 2: To solve.

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Use Cramer's rule to solve the systems in Exercises 22 through 24.
23. $| \begin{matrix} 5 x_{1} & - 3 x_{2} & = 1 \\ - 6 x_{1} & + 7 x_{2} & = 0 \end{matrix} |$