Solve Using a Matrix with Cramer's Rule 3x-4y-3z=-3 , -2y+3z=-9 , 3x+2z=26

$3 x - 4 y - 3 z = - 3$ , $- 2 y + 3 z = - 9$ , $3 x + 2 z = 26$

Represent the system of equations in matrix format.

$[\begin{array}{c} 3 & - 4 & - 3 \\ 0 & - 2 & 3 \\ 3 & 0 & 2 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 3 \\ - 9 \\ 26 \end{array}]$

Find the determinant of $[\begin{array}{c} 3 & - 4 & - 3 \\ 0 & - 2 & 3 \\ 3 & 0 & 2 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D = 3 | \begin{array}{c} - 2 & 3 \\ 0 & 2 \end{array} | + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 2 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

Find the determinant of $3 | \begin{array}{c} - 2 & 3 \\ 0 & 2 \end{array} |$ .

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

$D = 3 ((- 2) (2) + 0 \cdot 3) + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 2 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $2$ .

$D = 3 (- 4 + 0 \cdot 3) + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 2 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

Multiply $0$ by $3$ .

$D = 3 (- 4 + 0) + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 2 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

Simplify the expression.

Add $- 4$ and $0$ .

$D = 3 \cdot - 4 + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 2 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

Multiply $3$ by $- 4$ .

$D = - 12 + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 2 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D = - 12 + 0 + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & 3 \end{array} |$

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

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

$D = - 12 + 0 + 3 ((- 4) (3) + 2 \cdot - 3)$

Simplify the determinant.

Simplify each term.

Multiply $- 4$ by $3$ .

$D = - 12 + 0 + 3 (- 12 + 2 \cdot - 3)$

Multiply $2$ by $- 3$ .

$D = - 12 + 0 + 3 (- 12 - 6)$

Simplify the expression.

Subtract $6$ from $- 12$ .

$D = - 12 + 0 + 3 \cdot - 18$

Multiply $3$ by $- 18$ .

$D = - 12 + 0 - 54$

Add $- 12$ and $0$ .

$D = - 12 - 54$

Subtract $54$ from $- 12$ .

$D = - 66$

Find the determinant of $[\begin{array}{c} - 3 & - 4 & - 3 \\ - 9 & - 2 & 3 \\ 26 & 0 & 2 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{x} = 4 | \begin{array}{c} - 9 & 3 \\ 26 & 2 \end{array} | - 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Find the determinant of $4 | \begin{array}{c} - 9 & 3 \\ 26 & 2 \end{array} |$ .

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

$D_{x} = 4 ((- 9) (2) - 26 \cdot 3) - 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 9$ by $2$ .

$D_{x} = 4 (- 18 - 26 \cdot 3) - 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Multiply $- 26$ by $3$ .

$D_{x} = 4 (- 18 - 78) - 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Simplify the expression.

Subtract $78$ from $- 18$ .

$D_{x} = 4 \cdot - 96 - 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Multiply $4$ by $- 96$ .

$D_{x} = - 384 - 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Find the determinant of $- 2 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} |$ .

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

$D_{x} = - 384 - 2 ((- 3) (2) - 26 \cdot - 3) + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 3$ by $2$ .

$D_{x} = - 384 - 2 (- 6 - 26 \cdot - 3) + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Multiply $- 26$ by $- 3$ .

$D_{x} = - 384 - 2 (- 6 + 78) + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Simplify the expression.

Add $- 6$ and $78$ .

$D_{x} = - 384 - 2 \cdot 72 + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Multiply $- 2$ by $72$ .

$D_{x} = - 384 - 144 + 0 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{x} = - 384 - 144 + 0$

Subtract $144$ from $- 384$ .

$D_{x} = - 528 + 0$

Add $- 528$ and $0$ .

$D_{x} = - 528$

Find the determinant of $[\begin{array}{c} 3 & - 3 & - 3 \\ 0 & - 9 & 3 \\ 3 & 26 & 2 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{y} = 3 | \begin{array}{c} - 9 & 3 \\ 26 & 2 \end{array} | + 0 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Find the determinant of $3 | \begin{array}{c} - 9 & 3 \\ 26 & 2 \end{array} |$ .

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

$D_{y} = 3 ((- 9) (2) - 26 \cdot 3) + 0 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 9$ by $2$ .

$D_{y} = 3 (- 18 - 26 \cdot 3) + 0 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Multiply $- 26$ by $3$ .

$D_{y} = 3 (- 18 - 78) + 0 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Simplify the expression.

Subtract $78$ from $- 18$ .

$D_{y} = 3 \cdot - 96 + 0 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Multiply $3$ by $- 96$ .

$D_{y} = - 288 + 0 | \begin{array}{c} - 3 & - 3 \\ 26 & 2 \end{array} | + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{y} = - 288 + 0 + 3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$

Find the determinant of $3 | \begin{array}{c} - 3 & - 3 \\ - 9 & 3 \end{array} |$ .

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

$D_{y} = - 288 + 0 + 3 ((- 3) (3) + 9 \cdot - 3)$

Simplify the determinant.

Simplify each term.

Multiply $- 3$ by $3$ .

$D_{y} = - 288 + 0 + 3 (- 9 + 9 \cdot - 3)$

Multiply $9$ by $- 3$ .

$D_{y} = - 288 + 0 + 3 (- 9 - 27)$

Simplify the expression.

Subtract $27$ from $- 9$ .

$D_{y} = - 288 + 0 + 3 \cdot - 36$

Multiply $3$ by $- 36$ .

$D_{y} = - 288 + 0 - 108$

Add $- 288$ and $0$ .

$D_{y} = - 288 - 108$

Subtract $108$ from $- 288$ .

$D_{y} = - 396$

Find the determinant of $[\begin{array}{c} 3 & - 4 & - 3 \\ 0 & - 2 & - 9 \\ 3 & 0 & 26 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{z} = 3 | \begin{array}{c} - 2 & - 9 \\ 0 & 26 \end{array} | + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 26 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Find the determinant of $3 | \begin{array}{c} - 2 & - 9 \\ 0 & 26 \end{array} |$ .

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

$D_{z} = 3 ((- 2) (26) + 0 \cdot - 9) + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 26 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $26$ .

$D_{z} = 3 (- 52 + 0 \cdot - 9) + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 26 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Multiply $0$ by $- 9$ .

$D_{z} = 3 (- 52 + 0) + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 26 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Simplify the expression.

Add $- 52$ and $0$ .

$D_{z} = 3 \cdot - 52 + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 26 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Multiply $3$ by $- 52$ .

$D_{z} = - 156 + 0 | \begin{array}{c} - 4 & - 3 \\ 0 & 26 \end{array} | + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{z} = - 156 + 0 + 3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$

Find the determinant of $3 | \begin{array}{c} - 4 & - 3 \\ - 2 & - 9 \end{array} |$ .

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

$D_{z} = - 156 + 0 + 3 ((- 4) (- 9) + 2 \cdot - 3)$

Simplify the determinant.

Simplify each term.

Multiply $- 4$ by $- 9$ .

$D_{z} = - 156 + 0 + 3 (36 + 2 \cdot - 3)$

Multiply $2$ by $- 3$ .

$D_{z} = - 156 + 0 + 3 (36 - 6)$

Simplify the expression.

Subtract $6$ from $36$ .

$D_{z} = - 156 + 0 + 3 \cdot 30$

Multiply $3$ by $30$ .

$D_{z} = - 156 + 0 + 90$

Add $- 156$ and $0$ .

$D_{z} = - 156 + 90$

Add $- 156$ and $90$ .

$D_{z} = - 66$

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 528}{- 66}$ .

Remove parentheses.

$x = \frac{- 528}{- 66}$

Divide $- 528$ by $- 66$ .

$x = 8$

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 396}{- 66}$ .

Remove parentheses.

$y = \frac{- 396}{- 66}$

Divide $- 396$ by $- 66$ .

$y = 6$

Find the value of $z$ by Cramer\'s Rule, which states that $z = \frac{D_{z}}{D}$ . In this case, $z = \frac{- 66}{- 66}$ .

Remove parentheses.

$z = \frac{- 66}{- 66}$

Divide $- 66$ by $- 66$ .

$z = 1$

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

$x = 8, y = 6, z = 1$

Solve Using a Matrix with Cramer's Rule 3x-4y-3z=-3 , -2y+3z=-9 , 3x+2z=26

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