Solve Using a Matrix with Cramer's Rule 3x-2y+3z=5 , 3x+2y+4z=-4 , -5x+5y-4z=-2

$3 x - 2 y + 3 z = 5$ , $3 x + 2 y + 4 z = - 4$ , $- 5 x + 5 y - 4 z = - 2$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 3 & - 2 & 3 \\ 3 & 2 & 4 \\ - 5 & 5 & - 4 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 5 \\ - 4 \\ - 2 \end{array}]$

Step 2

Find the determinant of $[\begin{array}{c} 3 & - 2 & 3 \\ 3 & 2 & 4 \\ - 5 & 5 & - 4 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D = 3 | \begin{array}{c} 2 & 4 \\ 5 & - 4 \end{array} | - 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Find the determinant of $3 | \begin{array}{c} 2 & 4 \\ 5 & - 4 \end{array} |$ .

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

$D = 3 ((2) (- 4) - 5 \cdot 4) - 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $- 4$ .

$D = 3 (- 8 - 5 \cdot 4) - 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $- 5$ by $4$ .

$D = 3 (- 8 - 20) - 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the expression.

Subtract $20$ from $- 8$ .

$D = 3 \cdot - 28 - 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $3$ by $- 28$ .

$D = - 84 - 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Find the determinant of $- 3 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} |$ .

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

$D = - 84 - 3 ((- 2) (- 4) - 5 \cdot 3) - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $- 4$ .

$D = - 84 - 3 (8 - 5 \cdot 3) - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $- 5$ by $3$ .

$D = - 84 - 3 (8 - 15) - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the expression.

Subtract $15$ from $8$ .

$D = - 84 - 3 \cdot - 7 - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $- 3$ by $- 7$ .

$D = - 84 + 21 - 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Find the determinant of $- 5 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$ .

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

$D = - 84 + 21 - 5 ((- 2) (4) - 2 \cdot 3)$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $4$ .

$D = - 84 + 21 - 5 (- 8 - 2 \cdot 3)$

Multiply $- 2$ by $3$ .

$D = - 84 + 21 - 5 (- 8 - 6)$

Simplify the expression.

Subtract $6$ from $- 8$ .

$D = - 84 + 21 - 5 \cdot - 14$

Multiply $- 5$ by $- 14$ .

$D = - 84 + 21 + 70$

Add $- 84$ and $21$ .

$D = - 63 + 70$

Add $- 63$ and $70$ .

$D = 7$

Step 3

Find the determinant of $[\begin{array}{c} 5 & - 2 & 3 \\ - 4 & 2 & 4 \\ - 2 & 5 & - 4 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{x} = 5 | \begin{array}{c} 2 & 4 \\ 5 & - 4 \end{array} | + 4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Find the determinant of $5 | \begin{array}{c} 2 & 4 \\ 5 & - 4 \end{array} |$ .

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

$D_{x} = 5 ((2) (- 4) - 5 \cdot 4) + 4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $- 4$ .

$D_{x} = 5 (- 8 - 5 \cdot 4) + 4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $- 5$ by $4$ .

$D_{x} = 5 (- 8 - 20) + 4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the expression.

Subtract $20$ from $- 8$ .

$D_{x} = 5 \cdot - 28 + 4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $5$ by $- 28$ .

$D_{x} = - 140 + 4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} | - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Find the determinant of $4 | \begin{array}{c} - 2 & 3 \\ 5 & - 4 \end{array} |$ .

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

$D_{x} = - 140 + 4 ((- 2) (- 4) - 5 \cdot 3) - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $- 4$ .

$D_{x} = - 140 + 4 (8 - 5 \cdot 3) - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $- 5$ by $3$ .

$D_{x} = - 140 + 4 (8 - 15) - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Simplify the expression.

Subtract $15$ from $8$ .

$D_{x} = - 140 + 4 \cdot - 7 - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

Multiply $4$ by $- 7$ .

$D_{x} = - 140 - 28 - 2 | \begin{array}{c} - 2 & 3 \\ 2 & 4 \end{array} |$

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

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

$D_{x} = - 140 - 28 - 2 ((- 2) (4) - 2 \cdot 3)$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $4$ .

$D_{x} = - 140 - 28 - 2 (- 8 - 2 \cdot 3)$

Multiply $- 2$ by $3$ .

$D_{x} = - 140 - 28 - 2 (- 8 - 6)$

Simplify the expression.

Subtract $6$ from $- 8$ .

$D_{x} = - 140 - 28 - 2 \cdot - 14$

Multiply $- 2$ by $- 14$ .

$D_{x} = - 140 - 28 + 28$

Subtract $28$ from $- 140$ .

$D_{x} = - 168 + 28$

Add $- 168$ and $28$ .

$D_{x} = - 140$

Step 4

Find the determinant of $[\begin{array}{c} 3 & 5 & 3 \\ 3 & - 4 & 4 \\ - 5 & - 2 & - 4 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{y} = 3 | \begin{array}{c} - 4 & 4 \\ - 2 & - 4 \end{array} | - 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} | - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

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

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

$D_{y} = 3 ((- 4) (- 4) + 2 \cdot 4) - 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} | - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 4$ by $- 4$ .

$D_{y} = 3 (16 + 2 \cdot 4) - 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} | - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Multiply $2$ by $4$ .

$D_{y} = 3 (16 + 8) - 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} | - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Simplify the expression.

Add $16$ and $8$ .

$D_{y} = 3 \cdot 24 - 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} | - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Multiply $3$ by $24$ .

$D_{y} = 72 - 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} | - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Find the determinant of $- 3 | \begin{array}{c} 5 & 3 \\ - 2 & - 4 \end{array} |$ .

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

$D_{y} = 72 - 3 ((5) (- 4) + 2 \cdot 3) - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $- 4$ .

$D_{y} = 72 - 3 (- 20 + 2 \cdot 3) - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Multiply $2$ by $3$ .

$D_{y} = 72 - 3 (- 20 + 6) - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Simplify the expression.

Add $- 20$ and $6$ .

$D_{y} = 72 - 3 \cdot - 14 - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Multiply $- 3$ by $- 14$ .

$D_{y} = 72 + 42 - 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$

Find the determinant of $- 5 | \begin{array}{c} 5 & 3 \\ - 4 & 4 \end{array} |$ .

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

$D_{y} = 72 + 42 - 5 ((5) (4) + 4 \cdot 3)$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $4$ .

$D_{y} = 72 + 42 - 5 (20 + 4 \cdot 3)$

Multiply $4$ by $3$ .

$D_{y} = 72 + 42 - 5 (20 + 12)$

Simplify the expression.

Add $20$ and $12$ .

$D_{y} = 72 + 42 - 5 \cdot 32$

Multiply $- 5$ by $32$ .

$D_{y} = 72 + 42 - 160$

Add $72$ and $42$ .

$D_{y} = 114 - 160$

Subtract $160$ from $114$ .

$D_{y} = - 46$

Step 5

Find the determinant of $[\begin{array}{c} 3 & - 2 & 5 \\ 3 & 2 & - 4 \\ - 5 & 5 & - 2 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{z} = 3 | \begin{array}{c} 2 & - 4 \\ 5 & - 2 \end{array} | - 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} | - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Find the determinant of $3 | \begin{array}{c} 2 & - 4 \\ 5 & - 2 \end{array} |$ .

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

$D_{z} = 3 ((2) (- 2) - 5 \cdot - 4) - 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} | - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $- 2$ .

$D_{z} = 3 (- 4 - 5 \cdot - 4) - 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} | - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Multiply $- 5$ by $- 4$ .

$D_{z} = 3 (- 4 + 20) - 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} | - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Simplify the expression.

Add $- 4$ and $20$ .

$D_{z} = 3 \cdot 16 - 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} | - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Multiply $3$ by $16$ .

$D_{z} = 48 - 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} | - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Find the determinant of $- 3 | \begin{array}{c} - 2 & 5 \\ 5 & - 2 \end{array} |$ .

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

$D_{z} = 48 - 3 ((- 2) (- 2) - 5 \cdot 5) - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $- 2$ .

$D_{z} = 48 - 3 (4 - 5 \cdot 5) - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Multiply $- 5$ by $5$ .

$D_{z} = 48 - 3 (4 - 25) - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Simplify the expression.

Subtract $25$ from $4$ .

$D_{z} = 48 - 3 \cdot - 21 - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Multiply $- 3$ by $- 21$ .

$D_{z} = 48 + 63 - 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$

Find the determinant of $- 5 | \begin{array}{c} - 2 & 5 \\ 2 & - 4 \end{array} |$ .

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

$D_{z} = 48 + 63 - 5 ((- 2) (- 4) - 2 \cdot 5)$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $- 4$ .

$D_{z} = 48 + 63 - 5 (8 - 2 \cdot 5)$

Multiply $- 2$ by $5$ .

$D_{z} = 48 + 63 - 5 (8 - 10)$

Simplify the expression.

Subtract $10$ from $8$ .

$D_{z} = 48 + 63 - 5 \cdot - 2$

Multiply $- 5$ by $- 2$ .

$D_{z} = 48 + 63 + 10$

Add $48$ and $63$ .

$D_{z} = 111 + 10$

Add $111$ and $10$ .

$D_{z} = 121$

Step 6

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 140}{7}$ .

Remove parentheses.

$x = \frac{- 140}{7}$

Divide $- 140$ by $7$ .

$x = - 20$

Step 7

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 46}{7}$ .

Remove parentheses.

$y = \frac{- 46}{7}$

Move the negative in front of the fraction.

$y = - \frac{46}{7}$

Step 8

Remove parentheses.

$z = \frac{121}{7}$

Step 9

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

$x = - 20, y = - \frac{46}{7}, z = \frac{121}{7}$

Solve Using a Matrix with Cramer's Rule 3x-2y+3z=5 , 3x+2y+4z=-4 , -5x+5y-4z=-2

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