Finite Math Examples

$x + 4 y + 4 z = - 8$ , $x + 5 y + 4 z = - 10$ , $x + 4 y + 5 z = - 11$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 4 & 4 \\ 1 & 5 & 4 \\ 1 & 4 & 5 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 8 \\ - 10 \\ - 11 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 1 & 4 & 4 \\ 1 & 5 & 4 \\ 1 & 4 & 5 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 1 & 4 & 4 \\ 1 & 5 & 4 \\ 1 & 4 & 5 \end{array}]$ in determinant notation.

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

Step 2.2

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 row $1$ by its cofactor and add.

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Step 2.2.1

Consider the corresponding sign chart.

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

Step 2.2.2

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

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

Step 2.2.6

Multiply element $a_{12}$ by its cofactor.

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

Step 2.2.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

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

Step 2.2.8

Multiply element $a_{13}$ by its cofactor.

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

Step 2.2.9

Add the terms together.

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

Step 2.3

Evaluate $| \begin{array}{c} 5 & 4 \\ 4 & 5 \end{array} |$ .

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Step 2.3.1

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

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

Step 2.3.2

Simplify the determinant.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

Multiply $5$ by $5$ .

$1 (25 - 4 \cdot 4) - 4 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.3.2.1.2

Multiply $- 4$ by $4$ .

$1 (25 - 16) - 4 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.3.2.2

Subtract $16$ from $25$ .

$1 \cdot 9 - 4 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.4

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

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Step 2.4.1

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

$1 \cdot 9 - 4 (1 \cdot 5 - 1 \cdot 4) + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.4.2

Simplify the determinant.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Multiply $5$ by $1$ .

$1 \cdot 9 - 4 (5 - 1 \cdot 4) + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.4.2.1.2

Multiply $- 1$ by $4$ .

$1 \cdot 9 - 4 (5 - 4) + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.4.2.2

Subtract $4$ from $5$ .

$1 \cdot 9 - 4 \cdot 1 + 4 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 2.5

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

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Step 2.5.1

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

$1 \cdot 9 - 4 \cdot 1 + 4 (1 \cdot 4 - 1 \cdot 5)$

Step 2.5.2

Simplify the determinant.

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Step 2.5.2.1

Simplify each term.

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Step 2.5.2.1.1

Multiply $4$ by $1$ .

$1 \cdot 9 - 4 \cdot 1 + 4 (4 - 1 \cdot 5)$

Step 2.5.2.1.2

Multiply $- 1$ by $5$ .

$1 \cdot 9 - 4 \cdot 1 + 4 (4 - 5)$

Step 2.5.2.2

Subtract $5$ from $4$ .

$1 \cdot 9 - 4 \cdot 1 + 4 \cdot - 1$

Step 2.6

Simplify the determinant.

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Step 2.6.1

Simplify each term.

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Step 2.6.1.1

Multiply $9$ by $1$ .

$9 - 4 \cdot 1 + 4 \cdot - 1$

Step 2.6.1.2

Multiply $- 4$ by $1$ .

$9 - 4 + 4 \cdot - 1$

Step 2.6.1.3

Multiply $4$ by $- 1$ .

$9 - 4 - 4$

Step 2.6.2

Subtract $4$ from $9$ .

$5 - 4$

Step 2.6.3

Subtract $4$ from $5$ .

$1$

$D = 1$

Step 3

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

Step 4

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

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Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 8 \\ - 10 \\ - 11 \end{array}]$ .

$| \begin{array}{c} - 8 & 4 & 4 \\ - 10 & 5 & 4 \\ - 11 & 4 & 5 \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

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 $1$ by its cofactor and add.

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Step 4.2.1.1

Consider the corresponding sign chart.

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

Step 4.2.1.2

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

Step 4.2.1.3

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

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

Step 4.2.1.4

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

$- 8 | \begin{array}{c} 5 & 4 \\ 4 & 5 \end{array} |$

Step 4.2.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} |$

Step 4.2.1.6

Multiply element $a_{12}$ by its cofactor.

$- 4 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} |$

Step 4.2.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.1.8

Multiply element $a_{13}$ by its cofactor.

$4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.1.9

Add the terms together.

$- 8 | \begin{array}{c} 5 & 4 \\ 4 & 5 \end{array} | - 4 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} | + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.2

Evaluate $| \begin{array}{c} 5 & 4 \\ 4 & 5 \end{array} |$ .

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Step 4.2.2.1

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

$- 8 (5 \cdot 5 - 4 \cdot 4) - 4 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} | + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.2.2

Simplify the determinant.

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Step 4.2.2.2.1

Simplify each term.

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Step 4.2.2.2.1.1

Multiply $5$ by $5$ .

$- 8 (25 - 4 \cdot 4) - 4 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} | + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.2.2.1.2

Multiply $- 4$ by $4$ .

$- 8 (25 - 16) - 4 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} | + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.2.2.2

Subtract $16$ from $25$ .

$- 8 \cdot 9 - 4 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} | + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.3

Evaluate $| \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} |$ .

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Step 4.2.3.1

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

$- 8 \cdot 9 - 4 (- 10 \cdot 5 - (- 11 \cdot 4)) + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.3.2

Simplify the determinant.

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Step 4.2.3.2.1

Simplify each term.

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Step 4.2.3.2.1.1

Multiply $- 10$ by $5$ .

$- 8 \cdot 9 - 4 (- 50 - (- 11 \cdot 4)) + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.3.2.1.2

Multiply $- (- 11 \cdot 4)$ .

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Step 4.2.3.2.1.2.1

Multiply $- 11$ by $4$ .

$- 8 \cdot 9 - 4 (- 50 - - 44) + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.3.2.1.2.2

Multiply $- 1$ by $- 44$ .

$- 8 \cdot 9 - 4 (- 50 + 44) + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.3.2.2

Add $- 50$ and $44$ .

$- 8 \cdot 9 - 4 \cdot - 6 + 4 | \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$

Step 4.2.4

Evaluate $| \begin{array}{c} - 10 & 5 \\ - 11 & 4 \end{array} |$ .

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Step 4.2.4.1

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

$- 8 \cdot 9 - 4 \cdot - 6 + 4 (- 10 \cdot 4 - (- 11 \cdot 5))$

Step 4.2.4.2

Simplify the determinant.

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Step 4.2.4.2.1

Simplify each term.

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Step 4.2.4.2.1.1

Multiply $- 10$ by $4$ .

$- 8 \cdot 9 - 4 \cdot - 6 + 4 (- 40 - (- 11 \cdot 5))$

Step 4.2.4.2.1.2

Multiply $- (- 11 \cdot 5)$ .

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Step 4.2.4.2.1.2.1

Multiply $- 11$ by $5$ .

$- 8 \cdot 9 - 4 \cdot - 6 + 4 (- 40 - - 55)$

Step 4.2.4.2.1.2.2

Multiply $- 1$ by $- 55$ .

$- 8 \cdot 9 - 4 \cdot - 6 + 4 (- 40 + 55)$

Step 4.2.4.2.2

Add $- 40$ and $55$ .

$- 8 \cdot 9 - 4 \cdot - 6 + 4 \cdot 15$

Step 4.2.5

Simplify the determinant.

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Step 4.2.5.1

Simplify each term.

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Step 4.2.5.1.1

Multiply $- 8$ by $9$ .

$- 72 - 4 \cdot - 6 + 4 \cdot 15$

Step 4.2.5.1.2

Multiply $- 4$ by $- 6$ .

$- 72 + 24 + 4 \cdot 15$

Step 4.2.5.1.3

Multiply $4$ by $15$ .

$- 72 + 24 + 60$

Step 4.2.5.2

Add $- 72$ and $24$ .

$- 48 + 60$

Step 4.2.5.3

Add $- 48$ and $60$ .

$12$

$D_{x} = 12$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $1$ for $D$ and $12$ for $D_{x}$ in the formula.

$x = \frac{12}{1}$

Step 4.5

Divide $12$ by $1$ .

$x = 12$

Step 5

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

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Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} - 8 \\ - 10 \\ - 11 \end{array}]$ .

$| \begin{array}{c} 1 & - 8 & 4 \\ 1 & - 10 & 4 \\ 1 & - 11 & 5 \end{array} |$

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 row $1$ 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.

$| \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} |$

Step 5.2.1.4

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

$1 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} |$

Step 5.2.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

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

Step 5.2.1.6

Multiply element $a_{12}$ by its cofactor.

$8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} |$

Step 5.2.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.1.8

Multiply element $a_{13}$ by its cofactor.

$4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.1.9

Add the terms together.

$1 | \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} | + 8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.2

Evaluate $| \begin{array}{c} - 10 & 4 \\ - 11 & 5 \end{array} |$ .

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Step 5.2.2.1

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

$1 (- 10 \cdot 5 - (- 11 \cdot 4)) + 8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.2.2

Simplify the determinant.

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Step 5.2.2.2.1

Simplify each term.

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Step 5.2.2.2.1.1

Multiply $- 10$ by $5$ .

$1 (- 50 - (- 11 \cdot 4)) + 8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.2.2.1.2

Multiply $- (- 11 \cdot 4)$ .

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Step 5.2.2.2.1.2.1

Multiply $- 11$ by $4$ .

$1 (- 50 - - 44) + 8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.2.2.1.2.2

Multiply $- 1$ by $- 44$ .

$1 (- 50 + 44) + 8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.2.2.2

Add $- 50$ and $44$ .

$1 \cdot - 6 + 8 | \begin{array}{c} 1 & 4 \\ 1 & 5 \end{array} | + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.3

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

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Step 5.2.3.1

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

$1 \cdot - 6 + 8 (1 \cdot 5 - 1 \cdot 4) + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

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 $5$ by $1$ .

$1 \cdot - 6 + 8 (5 - 1 \cdot 4) + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.3.2.1.2

Multiply $- 1$ by $4$ .

$1 \cdot - 6 + 8 (5 - 4) + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.3.2.2

Subtract $4$ from $5$ .

$1 \cdot - 6 + 8 \cdot 1 + 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$ .

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Step 5.2.4.1

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

$1 \cdot - 6 + 8 \cdot 1 + 4 (1 \cdot - 11 - 1 \cdot - 10)$

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 $- 11$ by $1$ .

$1 \cdot - 6 + 8 \cdot 1 + 4 (- 11 - 1 \cdot - 10)$

Step 5.2.4.2.1.2

Multiply $- 1$ by $- 10$ .

$1 \cdot - 6 + 8 \cdot 1 + 4 (- 11 + 10)$

Step 5.2.4.2.2

Add $- 11$ and $10$ .

$1 \cdot - 6 + 8 \cdot 1 + 4 \cdot - 1$

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 $- 6$ by $1$ .

$- 6 + 8 \cdot 1 + 4 \cdot - 1$

Step 5.2.5.1.2

Multiply $8$ by $1$ .

$- 6 + 8 + 4 \cdot - 1$

Step 5.2.5.1.3

Multiply $4$ by $- 1$ .

$- 6 + 8 - 4$

Step 5.2.5.2

Add $- 6$ and $8$ .

$2 - 4$

Step 5.2.5.3

Subtract $4$ from $2$ .

$- 2$

$D_{y} = - 2$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $1$ for $D$ and $- 2$ for $D_{y}$ in the formula.

$y = \frac{- 2}{1}$

Step 5.5

Divide $- 2$ by $1$ .

$y = - 2$

Step 6

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

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Step 6.1

Replace column $3$ of the coefficient matrix that corresponds to the $z$ -coefficients of the system with $[\begin{array}{c} - 8 \\ - 10 \\ - 11 \end{array}]$ .

$| \begin{array}{c} 1 & 4 & - 8 \\ 1 & 5 & - 10 \\ 1 & 4 & - 11 \end{array} |$

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

$| \begin{array}{c} 5 & - 10 \\ 4 & - 11 \end{array} |$

Step 6.2.1.4

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

$1 | \begin{array}{c} 5 & - 10 \\ 4 & - 11 \end{array} |$

Step 6.2.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 6.2.1.6

Multiply element $a_{12}$ by its cofactor.

$- 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$

Step 6.2.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

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

Step 6.2.1.8

Multiply element $a_{13}$ by its cofactor.

$- 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.1.9

Add the terms together.

$1 | \begin{array}{c} 5 & - 10 \\ 4 & - 11 \end{array} | - 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} | - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.2

Evaluate $| \begin{array}{c} 5 & - 10 \\ 4 & - 11 \end{array} |$ .

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Step 6.2.2.1

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

$1 (5 \cdot - 11 - 4 \cdot - 10) - 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} | - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.2.2

Simplify the determinant.

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Step 6.2.2.2.1

Simplify each term.

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Step 6.2.2.2.1.1

Multiply $5$ by $- 11$ .

$1 (- 55 - 4 \cdot - 10) - 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} | - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.2.2.1.2

Multiply $- 4$ by $- 10$ .

$1 (- 55 + 40) - 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} | - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.2.2.2

Add $- 55$ and $40$ .

$1 \cdot - 15 - 4 | \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} | - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.3

Evaluate $| \begin{array}{c} 1 & - 10 \\ 1 & - 11 \end{array} |$ .

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Step 6.2.3.1

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

$1 \cdot - 15 - 4 (1 \cdot - 11 - 1 \cdot - 10) - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.3.2

Simplify the determinant.

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Step 6.2.3.2.1

Simplify each term.

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Step 6.2.3.2.1.1

Multiply $- 11$ by $1$ .

$1 \cdot - 15 - 4 (- 11 - 1 \cdot - 10) - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.3.2.1.2

Multiply $- 1$ by $- 10$ .

$1 \cdot - 15 - 4 (- 11 + 10) - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.3.2.2

Add $- 11$ and $10$ .

$1 \cdot - 15 - 4 \cdot - 1 - 8 | \begin{array}{c} 1 & 5 \\ 1 & 4 \end{array} |$

Step 6.2.4

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

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Step 6.2.4.1

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

$1 \cdot - 15 - 4 \cdot - 1 - 8 (1 \cdot 4 - 1 \cdot 5)$

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 $4$ by $1$ .

$1 \cdot - 15 - 4 \cdot - 1 - 8 (4 - 1 \cdot 5)$

Step 6.2.4.2.1.2

Multiply $- 1$ by $5$ .

$1 \cdot - 15 - 4 \cdot - 1 - 8 (4 - 5)$

Step 6.2.4.2.2

Subtract $5$ from $4$ .

$1 \cdot - 15 - 4 \cdot - 1 - 8 \cdot - 1$

Step 6.2.5

Simplify the determinant.

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Step 6.2.5.1

Simplify each term.

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Step 6.2.5.1.1

Multiply $- 15$ by $1$ .

$- 15 - 4 \cdot - 1 - 8 \cdot - 1$

Step 6.2.5.1.2

Multiply $- 4$ by $- 1$ .

$- 15 + 4 - 8 \cdot - 1$

Step 6.2.5.1.3

Multiply $- 8$ by $- 1$ .

$- 15 + 4 + 8$

Step 6.2.5.2

Add $- 15$ and $4$ .

$- 11 + 8$

Step 6.2.5.3

Add $- 11$ and $8$ .

$- 3$

$D_{z} = - 3$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $1$ for $D$ and $- 3$ for $D_{z}$ in the formula.

$z = \frac{- 3}{1}$

Step 6.5

Divide $- 3$ by $1$ .

$z = - 3$

Step 7

List the solution to the system of equations.

$x = 12$

$y = - 2$

$z = - 3$