Solve Using a Matrix with Cramer's Rule 4x-2y=-12 , 3x-y=-3

$4 x - 2 y = - 12$ , $3 x - y = - 3$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 4 & - 2 \\ 3 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 12 \\ - 3 \end{array}]$

Step 2

Find the determinant of $[\begin{array}{c} 4 & - 2 \\ 3 & - 1 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 4 & - 2 \\ 3 & - 1 \end{array}] = | \begin{array}{c} 4 & - 2 \\ 3 & - 1 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D = (4) (- 1) - 3 \cdot - 2$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $- 1$ .

$D = - 4 - 3 \cdot - 2$

Multiply $- 3$ by $- 2$ .

$D = - 4 + 6$

Add $- 4$ and $6$ .

$D = 2$

Step 3

Find the determinant of $[\begin{array}{c} - 12 & - 2 \\ - 3 & - 1 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} - 12 & - 2 \\ - 3 & - 1 \end{array}] = | \begin{array}{c} - 12 & - 2 \\ - 3 & - 1 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{x} = (- 12) (- 1) + 3 \cdot - 2$

Simplify the determinant.

Simplify each term.

Multiply $- 12$ by $- 1$ .

$D_{x} = 12 + 3 \cdot - 2$

Multiply $3$ by $- 2$ .

$D_{x} = 12 - 6$

Subtract $6$ from $12$ .

$D_{x} = 6$

Step 4

Find the determinant of $[\begin{array}{c} 4 & - 12 \\ 3 & - 3 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 4 & - 12 \\ 3 & - 3 \end{array}] = | \begin{array}{c} 4 & - 12 \\ 3 & - 3 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{y} = (4) (- 3) - 3 \cdot - 12$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $- 3$ .

$D_{y} = - 12 - 3 \cdot - 12$

Multiply $- 3$ by $- 12$ .

$D_{y} = - 12 + 36$

Add $- 12$ and $36$ .

$D_{y} = 24$

Step 5

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{6}{2}$ .

Remove parentheses.

$x = \frac{6}{2}$

Divide $6$ by $2$ .

$x = 3$

Step 6

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{24}{2}$ .

Remove parentheses.

$y = \frac{24}{2}$

Divide $24$ by $2$ .

$y = 12$

Step 7

The solution to the system of equations using Cramer\'s Rule.

$x = 3, y = 12$

Step 8

Solve Using a Matrix with Cramer's Rule 4x-2y=-12 , 3x-y=-3

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