Solve Using a Matrix with Cramer's Rule x+y=6 x-y=2

$x + y = 6$ $x - y = 2$

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 6 \\ 2 \end{array}]$

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

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

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

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

$D = (1) (- 1) - 1 \cdot 1$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $1$ .

$D = - 1 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$D = - 1 - 1$

Subtract $1$ from $- 1$ .

$D = - 2$

Find the determinant of $[\begin{array}{c} 6 & 1 \\ 2 & - 1 \end{array}]$ .

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

$determinant [\begin{array}{c} 6 & 1 \\ 2 & - 1 \end{array}] = | \begin{array}{c} 6 & 1 \\ 2 & - 1 \end{array} |$

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

$D_{x} = (6) (- 1) - 2 \cdot 1$

Simplify the determinant.

Simplify each term.

Multiply $6$ by $- 1$ .

$D_{x} = - 6 - 2 \cdot 1$

Multiply $- 2$ by $1$ .

$D_{x} = - 6 - 2$

Subtract $2$ from $- 6$ .

$D_{x} = - 8$

Find the determinant of $[\begin{array}{c} 1 & 6 \\ 1 & 2 \end{array}]$ .

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

$determinant [\begin{array}{c} 1 & 6 \\ 1 & 2 \end{array}] = | \begin{array}{c} 1 & 6 \\ 1 & 2 \end{array} |$

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

$D_{y} = (1) (2) - 1 \cdot 6$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $1$ .

$D_{y} = 2 - 1 \cdot 6$

Multiply $- 1$ by $6$ .

$D_{y} = 2 - 6$

Subtract $6$ from $2$ .

$D_{y} = - 4$

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 8}{- 2}$ .

Remove parentheses.

$x = \frac{- 8}{- 2}$

Divide $- 8$ by $- 2$ .

$x = 4$

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 4}{- 2}$ .

Remove parentheses.

$y = \frac{- 4}{- 2}$

Divide $- 4$ by $- 2$ .

$y = 2$

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

$x = 4, y = 2$

Solve Using a Matrix with Cramer's Rule x+y=6 x-y=2

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