Linear Algebra Examples

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

Step 1

Move all of the variables to the left side of each equation.

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Step 1.1

Move all terms containing variables to the left side of the equation.

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Step 1.1.1

Subtract $3 x$ from both sides of the equation.

$y - 3 x = z - 2$

$z = 3 x + 4$

$y = 5 z$

Step 1.1.2

Subtract $z$ from both sides of the equation.

$y - 3 x - z = - 2$

$z = 3 x + 4$

$y = 5 z$

$y - 3 x - z = - 2$

$z = 3 x + 4$

$y = 5 z$

Step 1.2

Reorder $y$ and $- 3 x$ .

$- 3 x + y - z = - 2$

$z = 3 x + 4$

$y = 5 z$

Step 1.3

Subtract $3 x$ from both sides of the equation.

$- 3 x + y - z = - 2$

$z - 3 x = 4$

$y = 5 z$

Step 1.4

Reorder $z$ and $- 3 x$ .

$- 3 x + y - z = - 2$

$- 3 x + z = 4$

$y = 5 z$

Step 1.5

Subtract $5 z$ from both sides of the equation.

$- 3 x + y - z = - 2$

$- 3 x + z = 4$

$y - 5 z = 0$

$- 3 x + y - z = - 2$

$- 3 x + z = 4$

$y - 5 z = 0$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}]$ in determinant notation.

$| \begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array} |$

Step 3.2

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in column $1$ by its cofactor and add.

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Step 3.2.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 3.2.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 3.2.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

Step 3.2.4

Multiply element $a_{11}$ by its cofactor.

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

Step 3.2.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

Step 3.2.6

Multiply element $a_{21}$ by its cofactor.

$3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

Step 3.2.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Step 3.2.8

Multiply element $a_{31}$ by its cofactor.

$0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Step 3.2.9

Add the terms together.

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Step 3.3

Multiply $0$ by $| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$ .

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 3.4

Evaluate $| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$ .

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Step 3.4.1

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

$- 3 (0 \cdot - 5 - 1 \cdot 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 3.4.2

Simplify the determinant.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply $0$ by $- 5$ .

$- 3 (0 - 1 \cdot 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 3.4.2.1.2

Multiply $- 1$ by $1$ .

$- 3 (0 - 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 3.4.2.2

Subtract $1$ from $0$ .

$- 3 \cdot - 1 + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 3.5

Evaluate $| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$ .

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Step 3.5.1

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

$- 3 \cdot - 1 + 3 (1 \cdot - 5 - 1 \cdot - 1) + 0$

Step 3.5.2

Simplify the determinant.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

Multiply $- 5$ by $1$ .

$- 3 \cdot - 1 + 3 (- 5 - 1 \cdot - 1) + 0$

Step 3.5.2.1.2

Multiply $- 1$ by $- 1$ .

$- 3 \cdot - 1 + 3 (- 5 + 1) + 0$

Step 3.5.2.2

Add $- 5$ and $1$ .

$- 3 \cdot - 1 + 3 \cdot - 4 + 0$

Step 3.6

Simplify the determinant.

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Step 3.6.1

Simplify each term.

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Step 3.6.1.1

Multiply $- 3$ by $- 1$ .

$3 + 3 \cdot - 4 + 0$

Step 3.6.1.2

Multiply $3$ by $- 4$ .

$3 - 12 + 0$

Step 3.6.2

Subtract $12$ from $3$ .

$- 9 + 0$

Step 3.6.3

Add $- 9$ and $0$ .

$- 9$

$D = - 9$

Step 4

Since the determinant is not $0$ , the system can be solved using Cramer's Rule.

Step 5

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

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Step 5.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 2 & 1 & - 1 \\ 4 & 0 & 1 \\ 0 & 1 & - 5 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in column $1$ by its cofactor and add.

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Step 5.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.2.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.2.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

Step 5.2.1.4

Multiply element $a_{11}$ by its cofactor.

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

Step 5.2.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

Step 5.2.1.6

Multiply element $a_{21}$ by its cofactor.

$- 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

Step 5.2.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Step 5.2.1.8

Multiply element $a_{31}$ by its cofactor.

$0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Step 5.2.1.9

Add the terms together.

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Step 5.2.2

Multiply $0$ by $| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$ .

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 5.2.3

Evaluate $| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$ .

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Step 5.2.3.1

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

$- 2 (0 \cdot - 5 - 1 \cdot 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 5.2.3.2

Simplify the determinant.

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Step 5.2.3.2.1

Simplify each term.

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Step 5.2.3.2.1.1

Multiply $0$ by $- 5$ .

$- 2 (0 - 1 \cdot 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 5.2.3.2.1.2

Multiply $- 1$ by $1$ .

$- 2 (0 - 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 5.2.3.2.2

Subtract $1$ from $0$ .

$- 2 \cdot - 1 - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

Step 5.2.4

Evaluate $| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$ .

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Step 5.2.4.1

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

$- 2 \cdot - 1 - 4 (1 \cdot - 5 - 1 \cdot - 1) + 0$

Step 5.2.4.2

Simplify the determinant.

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Step 5.2.4.2.1

Simplify each term.

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Step 5.2.4.2.1.1

Multiply $- 5$ by $1$ .

$- 2 \cdot - 1 - 4 (- 5 - 1 \cdot - 1) + 0$

Step 5.2.4.2.1.2

Multiply $- 1$ by $- 1$ .

$- 2 \cdot - 1 - 4 (- 5 + 1) + 0$

Step 5.2.4.2.2

Add $- 5$ and $1$ .

$- 2 \cdot - 1 - 4 \cdot - 4 + 0$

Step 5.2.5

Simplify the determinant.

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Step 5.2.5.1

Simplify each term.

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Step 5.2.5.1.1

Multiply $- 2$ by $- 1$ .

$2 - 4 \cdot - 4 + 0$

Step 5.2.5.1.2

Multiply $- 4$ by $- 4$ .

$2 + 16 + 0$

Step 5.2.5.2

Add $2$ and $16$ .

$18 + 0$

Step 5.2.5.3

Add $18$ and $0$ .

$18$

$D_{x} = 18$

Step 5.3

Use the formula to solve for $x$ .

$x = \frac{D_{x}}{D}$

Step 5.4

Substitute $- 9$ for $D$ and $18$ for $D_{x}$ in the formula.

$x = \frac{18}{- 9}$

Step 5.5

Divide $18$ by $- 9$ .

$x = - 2$

Step 6

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

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Step 6.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 3 & - 2 & - 1 \\ - 3 & 4 & 1 \\ 0 & 0 & - 5 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $3$ by its cofactor and add.

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Step 6.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 6.2.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 6.2.1.3

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Step 6.2.1.4

Multiply element $a_{31}$ by its cofactor.

$0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Step 6.2.1.5

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

$| \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} |$

Step 6.2.1.6

Multiply element $a_{32}$ by its cofactor.

$0 | \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} |$

Step 6.2.1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

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

Step 6.2.1.8

Multiply element $a_{33}$ by its cofactor.

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

Step 6.2.1.9

Add the terms together.

$0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} | - 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

Step 6.2.2

Multiply $0$ by $| \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$ .

$0 + 0 | \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} | - 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

Step 6.2.3

Multiply $0$ by $| \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} |$ .

$0 + 0 - 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

Step 6.2.4

Evaluate $| \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$ .

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Step 6.2.4.1

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

$0 + 0 - 5 (- 3 \cdot 4 - (- 3 \cdot - 2))$

Step 6.2.4.2

Simplify the determinant.

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Step 6.2.4.2.1

Simplify each term.

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Step 6.2.4.2.1.1

Multiply $- 3$ by $4$ .

$0 + 0 - 5 (- 12 - (- 3 \cdot - 2))$

Step 6.2.4.2.1.2

Multiply $- (- 3 \cdot - 2)$ .

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Step 6.2.4.2.1.2.1

Multiply $- 3$ by $- 2$ .

$0 + 0 - 5 (- 12 - 1 \cdot 6)$

Step 6.2.4.2.1.2.2

Multiply $- 1$ by $6$ .

$0 + 0 - 5 (- 12 - 6)$

Step 6.2.4.2.2

Subtract $6$ from $- 12$ .

$0 + 0 - 5 \cdot - 18$

Step 6.2.5

Simplify the determinant.

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Step 6.2.5.1

Multiply $- 5$ by $- 18$ .

$0 + 0 + 90$

Step 6.2.5.2

Add $0$ and $0$ .

$0 + 90$

Step 6.2.5.3

Add $0$ and $90$ .

$90$

$D_{y} = 90$

Step 6.3

Use the formula to solve for $y$ .

$y = \frac{D_{y}}{D}$

Step 6.4

Substitute $- 9$ for $D$ and $90$ for $D_{y}$ in the formula.

$y = \frac{90}{- 9}$

Step 6.5

Divide $90$ by $- 9$ .

$y = - 10$

Step 7

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

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Step 7.1

Replace column $3$ of the coefficient matrix that corresponds to the $z$ -coefficients of the system with $[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 3 & 1 & - 2 \\ - 3 & 0 & 4 \\ 0 & 1 & 0 \end{array} |$

Step 7.2

Find the determinant.

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Step 7.2.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $3$ by its cofactor and add.

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Step 7.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 7.2.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 7.2.1.3

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Step 7.2.1.4

Multiply element $a_{31}$ by its cofactor.

$0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Step 7.2.1.5

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 7.2.1.6

Multiply element $a_{32}$ by its cofactor.

$- 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

Step 7.2.1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

$| \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

Step 7.2.1.8

Multiply element $a_{33}$ by its cofactor.

$0 | \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

Step 7.2.1.9

Add the terms together.

$0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} | - 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} | + 0 | \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

Step 7.2.2

Multiply $0$ by $| \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$ .

$0 - 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} | + 0 | \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

Step 7.2.3

Multiply $0$ by $| \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$ .

$0 - 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} | + 0$

Step 7.2.4

Evaluate $| \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$ .

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Step 7.2.4.1

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

$0 - 1 (- 3 \cdot 4 - (- 3 \cdot - 2)) + 0$

Step 7.2.4.2

Simplify the determinant.

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Step 7.2.4.2.1

Simplify each term.

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Step 7.2.4.2.1.1

Multiply $- 3$ by $4$ .

$0 - 1 (- 12 - (- 3 \cdot - 2)) + 0$

Step 7.2.4.2.1.2

Multiply $- (- 3 \cdot - 2)$ .

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Step 7.2.4.2.1.2.1

Multiply $- 3$ by $- 2$ .

$0 - 1 (- 12 - 1 \cdot 6) + 0$

Step 7.2.4.2.1.2.2

Multiply $- 1$ by $6$ .

$0 - 1 (- 12 - 6) + 0$

Step 7.2.4.2.2

Subtract $6$ from $- 12$ .

$0 - 1 \cdot - 18 + 0$

Step 7.2.5

Simplify the determinant.

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Step 7.2.5.1

Multiply $- 1$ by $- 18$ .

$0 + 18 + 0$

Step 7.2.5.2

Add $0$ and $18$ .

$18 + 0$

Step 7.2.5.3

Add $18$ and $0$ .

$18$

$D_{z} = 18$

Step 7.3

Use the formula to solve for $z$ .

$z = \frac{D_{z}}{D}$

Step 7.4

Substitute $- 9$ for $D$ and $18$ for $D_{z}$ in the formula.

$z = \frac{18}{- 9}$

Step 7.5

Divide $18$ by $- 9$ .

$z = - 2$

Step 8

List the solution to the system of equations.

$x = - 2$

$y = - 10$

$z = - 2$

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