Solve Using a Matrix with Cramer's Rule -4x-4y+2z=4 , -2y+3z=22 , 2x+3z=28

$- 4 x - 4 y + 2 z = 4$ , $- 2 y + 3 z = 22$ , $2 x + 3 z = 28$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} - 4 & - 4 & 2 \\ 0 & - 2 & 3 \\ 2 & 0 & 3 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ 22 \\ 28 \end{array}]$

Step 2

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

Set up the determinant by breaking it into smaller components.

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

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $3$ .

$D = - 4 (- 6 + 0 \cdot 3) + 0 | \begin{array}{c} - 4 & 2 \\ 0 & 3 \end{array} | + 2 | \begin{array}{c} - 4 & 2 \\ - 2 & 3 \end{array} |$

Multiply $0$ by $3$ .

$D = - 4 (- 6 + 0) + 0 | \begin{array}{c} - 4 & 2 \\ 0 & 3 \end{array} | + 2 | \begin{array}{c} - 4 & 2 \\ - 2 & 3 \end{array} |$

Simplify the expression.

Add $- 6$ and $0$ .

$D = - 4 \cdot - 6 + 0 | \begin{array}{c} - 4 & 2 \\ 0 & 3 \end{array} | + 2 | \begin{array}{c} - 4 & 2 \\ - 2 & 3 \end{array} |$

Multiply $- 4$ by $- 6$ .

$D = 24 + 0 | \begin{array}{c} - 4 & 2 \\ 0 & 3 \end{array} | + 2 | \begin{array}{c} - 4 & 2 \\ - 2 & 3 \end{array} |$

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

$D = 24 + 0 + 2 | \begin{array}{c} - 4 & 2 \\ - 2 & 3 \end{array} |$

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

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

$D = 24 + 0 + 2 ((- 4) (3) + 2 \cdot 2)$

Simplify the determinant.

Simplify each term.

Multiply $- 4$ by $3$ .

$D = 24 + 0 + 2 (- 12 + 2 \cdot 2)$

Multiply $2$ by $2$ .

$D = 24 + 0 + 2 (- 12 + 4)$

Simplify the expression.

Add $- 12$ and $4$ .

$D = 24 + 0 + 2 \cdot - 8$

Multiply $2$ by $- 8$ .

$D = 24 + 0 - 16$

Add $24$ and $0$ .

$D = 24 - 16$

Subtract $16$ from $24$ .

$D = 8$

Step 3

Find the determinant of $[\begin{array}{c} 4 & - 4 & 2 \\ 22 & - 2 & 3 \\ 28 & 0 & 3 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{x} = 4 | \begin{array}{c} 22 & 3 \\ 28 & 3 \end{array} | - 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Find the determinant of $4 | \begin{array}{c} 22 & 3 \\ 28 & 3 \end{array} |$ .

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

$D_{x} = 4 ((22) (3) - 28 \cdot 3) - 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $22$ by $3$ .

$D_{x} = 4 (66 - 28 \cdot 3) - 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Multiply $- 28$ by $3$ .

$D_{x} = 4 (66 - 84) - 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Simplify the expression.

Subtract $84$ from $66$ .

$D_{x} = 4 \cdot - 18 - 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Multiply $4$ by $- 18$ .

$D_{x} = - 72 - 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Find the determinant of $- 2 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} |$ .

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

$D_{x} = - 72 - 2 ((4) (3) - 28 \cdot 2) + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $3$ .

$D_{x} = - 72 - 2 (12 - 28 \cdot 2) + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Multiply $- 28$ by $2$ .

$D_{x} = - 72 - 2 (12 - 56) + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Simplify the expression.

Subtract $56$ from $12$ .

$D_{x} = - 72 - 2 \cdot - 44 + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Multiply $- 2$ by $- 44$ .

$D_{x} = - 72 + 88 + 0 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

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

$D_{x} = - 72 + 88 + 0$

Add $- 72$ and $88$ .

$D_{x} = 16 + 0$

Add $16$ and $0$ .

$D_{x} = 16$

Step 4

Find the determinant of $[\begin{array}{c} - 4 & 4 & 2 \\ 0 & 22 & 3 \\ 2 & 28 & 3 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{y} = - 4 | \begin{array}{c} 22 & 3 \\ 28 & 3 \end{array} | + 0 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Find the determinant of $- 4 | \begin{array}{c} 22 & 3 \\ 28 & 3 \end{array} |$ .

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

$D_{y} = - 4 ((22) (3) - 28 \cdot 3) + 0 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $22$ by $3$ .

$D_{y} = - 4 (66 - 28 \cdot 3) + 0 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Multiply $- 28$ by $3$ .

$D_{y} = - 4 (66 - 84) + 0 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Simplify the expression.

Subtract $84$ from $66$ .

$D_{y} = - 4 \cdot - 18 + 0 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Multiply $- 4$ by $- 18$ .

$D_{y} = 72 + 0 | \begin{array}{c} 4 & 2 \\ 28 & 3 \end{array} | + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

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

$D_{y} = 72 + 0 + 2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$

Find the determinant of $2 | \begin{array}{c} 4 & 2 \\ 22 & 3 \end{array} |$ .

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

$D_{y} = 72 + 0 + 2 ((4) (3) - 22 \cdot 2)$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $3$ .

$D_{y} = 72 + 0 + 2 (12 - 22 \cdot 2)$

Multiply $- 22$ by $2$ .

$D_{y} = 72 + 0 + 2 (12 - 44)$

Simplify the expression.

Subtract $44$ from $12$ .

$D_{y} = 72 + 0 + 2 \cdot - 32$

Multiply $2$ by $- 32$ .

$D_{y} = 72 + 0 - 64$

Add $72$ and $0$ .

$D_{y} = 72 - 64$

Subtract $64$ from $72$ .

$D_{y} = 8$

Step 5

Find the determinant of $[\begin{array}{c} - 4 & - 4 & 4 \\ 0 & - 2 & 22 \\ 2 & 0 & 28 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{z} = - 4 | \begin{array}{c} - 2 & 22 \\ 0 & 28 \end{array} | + 0 | \begin{array}{c} - 4 & 4 \\ 0 & 28 \end{array} | + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

Find the determinant of $- 4 | \begin{array}{c} - 2 & 22 \\ 0 & 28 \end{array} |$ .

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

$D_{z} = - 4 ((- 2) (28) + 0 \cdot 22) + 0 | \begin{array}{c} - 4 & 4 \\ 0 & 28 \end{array} | + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $28$ .

$D_{z} = - 4 (- 56 + 0 \cdot 22) + 0 | \begin{array}{c} - 4 & 4 \\ 0 & 28 \end{array} | + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

Multiply $0$ by $22$ .

$D_{z} = - 4 (- 56 + 0) + 0 | \begin{array}{c} - 4 & 4 \\ 0 & 28 \end{array} | + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

Simplify the expression.

Add $- 56$ and $0$ .

$D_{z} = - 4 \cdot - 56 + 0 | \begin{array}{c} - 4 & 4 \\ 0 & 28 \end{array} | + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

Multiply $- 4$ by $- 56$ .

$D_{z} = 224 + 0 | \begin{array}{c} - 4 & 4 \\ 0 & 28 \end{array} | + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

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

$D_{z} = 224 + 0 + 2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$

Find the determinant of $2 | \begin{array}{c} - 4 & 4 \\ - 2 & 22 \end{array} |$ .

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

$D_{z} = 224 + 0 + 2 ((- 4) (22) + 2 \cdot 4)$

Simplify the determinant.

Simplify each term.

Multiply $- 4$ by $22$ .

$D_{z} = 224 + 0 + 2 (- 88 + 2 \cdot 4)$

Multiply $2$ by $4$ .

$D_{z} = 224 + 0 + 2 (- 88 + 8)$

Simplify the expression.

Add $- 88$ and $8$ .

$D_{z} = 224 + 0 + 2 \cdot - 80$

Multiply $2$ by $- 80$ .

$D_{z} = 224 + 0 - 160$

Add $224$ and $0$ .

$D_{z} = 224 - 160$

Subtract $160$ from $224$ .

$D_{z} = 64$

Step 6

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{16}{8}$ .

Remove parentheses.

$x = \frac{16}{8}$

Divide $16$ by $8$ .

$x = 2$

Step 7

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

Remove parentheses.

$y = \frac{8}{8}$

Divide $8$ by $8$ .

$y = 1$

Step 8

Find the value of $z$ by Cramer\'s Rule, which states that $z = \frac{D_{z}}{D}$ . In this case, $z = \frac{64}{8}$ .

Remove parentheses.

$z = \frac{64}{8}$

Divide $64$ by $8$ .

$z = 8$

Step 9

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

$x = 2, y = 1, z = 8$

Solve Using a Matrix with Cramer's Rule -4x-4y+2z=4 , -2y+3z=22 , 2x+3z=28

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