Precalculus Examples

$4 x - 3 y + 2 z = 15$ , $- 2 x + y - 3 z = - 15$ , $5 x - 5 y + 2 z = 18$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 4 & - 3 & 2 \\ - 2 & 1 & - 3 \\ 5 & - 5 & 2 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 15 \\ - 15 \\ 18 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 4 & - 3 & 2 \\ - 2 & 1 & - 3 \\ 5 & - 5 & 2 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 4 & - 3 & 2 \\ - 2 & 1 & - 3 \\ 5 & - 5 & 2 \end{array}]$ in determinant notation.

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

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Add the terms together.

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

Step 2.3

Evaluate $| \begin{array}{c} 1 & - 3 \\ - 5 & 2 \end{array} |$ .

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Step 2.3.1

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

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

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 $2$ by $1$ .

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

Step 2.3.2.1.2

Multiply $- (- 5 \cdot - 3)$ .

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Step 2.3.2.1.2.1

Multiply $- 5$ by $- 3$ .

$4 (2 - 1 \cdot 15) + 3 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 2.3.2.1.2.2

Multiply $- 1$ by $15$ .

$4 (2 - 15) + 3 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 2.3.2.2

Subtract $15$ from $2$ .

$4 \cdot - 13 + 3 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 2.4

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

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Step 2.4.1

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

$4 \cdot - 13 + 3 (- 2 \cdot 2 - 5 \cdot - 3) + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

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 $- 2$ by $2$ .

$4 \cdot - 13 + 3 (- 4 - 5 \cdot - 3) + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 2.4.2.1.2

Multiply $- 5$ by $- 3$ .

$4 \cdot - 13 + 3 (- 4 + 15) + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 2.4.2.2

Add $- 4$ and $15$ .

$4 \cdot - 13 + 3 \cdot 11 + 2 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 2.5

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

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Step 2.5.1

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

$4 \cdot - 13 + 3 \cdot 11 + 2 (- 2 \cdot - 5 - 5 \cdot 1)$

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 $- 2$ by $- 5$ .

$4 \cdot - 13 + 3 \cdot 11 + 2 (10 - 5 \cdot 1)$

Step 2.5.2.1.2

Multiply $- 5$ by $1$ .

$4 \cdot - 13 + 3 \cdot 11 + 2 (10 - 5)$

Step 2.5.2.2

Subtract $5$ from $10$ .

$4 \cdot - 13 + 3 \cdot 11 + 2 \cdot 5$

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 $4$ by $- 13$ .

$- 52 + 3 \cdot 11 + 2 \cdot 5$

Step 2.6.1.2

Multiply $3$ by $11$ .

$- 52 + 33 + 2 \cdot 5$

Step 2.6.1.3

Multiply $2$ by $5$ .

$- 52 + 33 + 10$

Step 2.6.2

Add $- 52$ and $33$ .

$- 19 + 10$

Step 2.6.3

Add $- 19$ and $10$ .

$- 9$

$D = - 9$

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 $[\begin{array}{c} 15 \\ - 15 \\ 18 \end{array}]$ .

$| \begin{array}{c} 15 & - 3 & 2 \\ - 15 & 1 & - 3 \\ 18 & - 5 & 2 \end{array} |$

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.

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

Step 4.2.1.4

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

$15 | \begin{array}{c} 1 & - 3 \\ - 5 & 2 \end{array} |$

Step 4.2.1.5

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

$| \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} |$

Step 4.2.1.6

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

$3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} |$

Step 4.2.1.7

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

$| \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.1.8

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

$2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.1.9

Add the terms together.

$15 | \begin{array}{c} 1 & - 3 \\ - 5 & 2 \end{array} | + 3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.2

Evaluate $| \begin{array}{c} 1 & - 3 \\ - 5 & 2 \end{array} |$ .

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Step 4.2.2.1

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

$15 (1 \cdot 2 - (- 5 \cdot - 3)) + 3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

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 $2$ by $1$ .

$15 (2 - (- 5 \cdot - 3)) + 3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.2.2.1.2

Multiply $- (- 5 \cdot - 3)$ .

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Step 4.2.2.2.1.2.1

Multiply $- 5$ by $- 3$ .

$15 (2 - 1 \cdot 15) + 3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.2.2.1.2.2

Multiply $- 1$ by $15$ .

$15 (2 - 15) + 3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.2.2.2

Subtract $15$ from $2$ .

$15 \cdot - 13 + 3 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.3

Evaluate $| \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} |$ .

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Step 4.2.3.1

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

$15 \cdot - 13 + 3 (- 15 \cdot 2 - 18 \cdot - 3) + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

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 $- 15$ by $2$ .

$15 \cdot - 13 + 3 (- 30 - 18 \cdot - 3) + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.3.2.1.2

Multiply $- 18$ by $- 3$ .

$15 \cdot - 13 + 3 (- 30 + 54) + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.3.2.2

Add $- 30$ and $54$ .

$15 \cdot - 13 + 3 \cdot 24 + 2 | \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$

Step 4.2.4

Evaluate $| \begin{array}{c} - 15 & 1 \\ 18 & - 5 \end{array} |$ .

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Step 4.2.4.1

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

$15 \cdot - 13 + 3 \cdot 24 + 2 (- 15 \cdot - 5 - 18 \cdot 1)$

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 $- 15$ by $- 5$ .

$15 \cdot - 13 + 3 \cdot 24 + 2 (75 - 18 \cdot 1)$

Step 4.2.4.2.1.2

Multiply $- 18$ by $1$ .

$15 \cdot - 13 + 3 \cdot 24 + 2 (75 - 18)$

Step 4.2.4.2.2

Subtract $18$ from $75$ .

$15 \cdot - 13 + 3 \cdot 24 + 2 \cdot 57$

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 $15$ by $- 13$ .

$- 195 + 3 \cdot 24 + 2 \cdot 57$

Step 4.2.5.1.2

Multiply $3$ by $24$ .

$- 195 + 72 + 2 \cdot 57$

Step 4.2.5.1.3

Multiply $2$ by $57$ .

$- 195 + 72 + 114$

Step 4.2.5.2

Add $- 195$ and $72$ .

$- 123 + 114$

Step 4.2.5.3

Add $- 123$ and $114$ .

$- 9$

$D_{x} = - 9$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

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

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

Step 4.5

Divide $- 9$ by $- 9$ .

$x = 1$

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 $[\begin{array}{c} 15 \\ - 15 \\ 18 \end{array}]$ .

$| \begin{array}{c} 4 & 15 & 2 \\ - 2 & - 15 & - 3 \\ 5 & 18 & 2 \end{array} |$

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.

$| \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} |$

Step 5.2.1.4

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

$4 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} |$

Step 5.2.1.5

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

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

Step 5.2.1.6

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

$- 15 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} |$

Step 5.2.1.7

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

$| \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.1.8

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

$2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.1.9

Add the terms together.

$4 | \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} | - 15 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.2

Evaluate $| \begin{array}{c} - 15 & - 3 \\ 18 & 2 \end{array} |$ .

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Step 5.2.2.1

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

$4 (- 15 \cdot 2 - 18 \cdot - 3) - 15 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

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 $- 15$ by $2$ .

$4 (- 30 - 18 \cdot - 3) - 15 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.2.2.1.2

Multiply $- 18$ by $- 3$ .

$4 (- 30 + 54) - 15 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.2.2.2

Add $- 30$ and $54$ .

$4 \cdot 24 - 15 | \begin{array}{c} - 2 & - 3 \\ 5 & 2 \end{array} | + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.3

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

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Step 5.2.3.1

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

$4 \cdot 24 - 15 (- 2 \cdot 2 - 5 \cdot - 3) + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

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 $- 2$ by $2$ .

$4 \cdot 24 - 15 (- 4 - 5 \cdot - 3) + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.3.2.1.2

Multiply $- 5$ by $- 3$ .

$4 \cdot 24 - 15 (- 4 + 15) + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.3.2.2

Add $- 4$ and $15$ .

$4 \cdot 24 - 15 \cdot 11 + 2 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$ .

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Step 5.2.4.1

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

$4 \cdot 24 - 15 \cdot 11 + 2 (- 2 \cdot 18 - 5 \cdot - 15)$

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 $- 2$ by $18$ .

$4 \cdot 24 - 15 \cdot 11 + 2 (- 36 - 5 \cdot - 15)$

Step 5.2.4.2.1.2

Multiply $- 5$ by $- 15$ .

$4 \cdot 24 - 15 \cdot 11 + 2 (- 36 + 75)$

Step 5.2.4.2.2

Add $- 36$ and $75$ .

$4 \cdot 24 - 15 \cdot 11 + 2 \cdot 39$

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 $4$ by $24$ .

$96 - 15 \cdot 11 + 2 \cdot 39$

Step 5.2.5.1.2

Multiply $- 15$ by $11$ .

$96 - 165 + 2 \cdot 39$

Step 5.2.5.1.3

Multiply $2$ by $39$ .

$96 - 165 + 78$

Step 5.2.5.2

Subtract $165$ from $96$ .

$- 69 + 78$

Step 5.2.5.3

Add $- 69$ and $78$ .

$9$

$D_{y} = 9$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

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

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

Step 5.5

Divide $9$ by $- 9$ .

$y = - 1$

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 $[\begin{array}{c} 15 \\ - 15 \\ 18 \end{array}]$ .

$| \begin{array}{c} 4 & - 3 & 15 \\ - 2 & 1 & - 15 \\ 5 & - 5 & 18 \end{array} |$

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.

$| \begin{array}{c} 1 & - 15 \\ - 5 & 18 \end{array} |$

Step 6.2.1.4

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

$4 | \begin{array}{c} 1 & - 15 \\ - 5 & 18 \end{array} |$

Step 6.2.1.5

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

$| \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 6.2.1.6

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

$3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$

Step 6.2.1.7

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

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

Step 6.2.1.8

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

$15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.1.9

Add the terms together.

$4 | \begin{array}{c} 1 & - 15 \\ - 5 & 18 \end{array} | + 3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} | + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.2

Evaluate $| \begin{array}{c} 1 & - 15 \\ - 5 & 18 \end{array} |$ .

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Step 6.2.2.1

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

$4 (1 \cdot 18 - (- 5 \cdot - 15)) + 3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} | + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

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 $18$ by $1$ .

$4 (18 - (- 5 \cdot - 15)) + 3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} | + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.2.2.1.2

Multiply $- (- 5 \cdot - 15)$ .

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Step 6.2.2.2.1.2.1

Multiply $- 5$ by $- 15$ .

$4 (18 - 1 \cdot 75) + 3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} | + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.2.2.1.2.2

Multiply $- 1$ by $75$ .

$4 (18 - 75) + 3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} | + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.2.2.2

Subtract $75$ from $18$ .

$4 \cdot - 57 + 3 | \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} | + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.3

Evaluate $| \begin{array}{c} - 2 & - 15 \\ 5 & 18 \end{array} |$ .

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Step 6.2.3.1

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

$4 \cdot - 57 + 3 (- 2 \cdot 18 - 5 \cdot - 15) + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

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 $- 2$ by $18$ .

$4 \cdot - 57 + 3 (- 36 - 5 \cdot - 15) + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.3.2.1.2

Multiply $- 5$ by $- 15$ .

$4 \cdot - 57 + 3 (- 36 + 75) + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.3.2.2

Add $- 36$ and $75$ .

$4 \cdot - 57 + 3 \cdot 39 + 15 | \begin{array}{c} - 2 & 1 \\ 5 & - 5 \end{array} |$

Step 6.2.4

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

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Step 6.2.4.1

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

$4 \cdot - 57 + 3 \cdot 39 + 15 (- 2 \cdot - 5 - 5 \cdot 1)$

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 $- 2$ by $- 5$ .

$4 \cdot - 57 + 3 \cdot 39 + 15 (10 - 5 \cdot 1)$

Step 6.2.4.2.1.2

Multiply $- 5$ by $1$ .

$4 \cdot - 57 + 3 \cdot 39 + 15 (10 - 5)$

Step 6.2.4.2.2

Subtract $5$ from $10$ .

$4 \cdot - 57 + 3 \cdot 39 + 15 \cdot 5$

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 $4$ by $- 57$ .

$- 228 + 3 \cdot 39 + 15 \cdot 5$

Step 6.2.5.1.2

Multiply $3$ by $39$ .

$- 228 + 117 + 15 \cdot 5$

Step 6.2.5.1.3

Multiply $15$ by $5$ .

$- 228 + 117 + 75$

Step 6.2.5.2

Add $- 228$ and $117$ .

$- 111 + 75$

Step 6.2.5.3

Add $- 111$ and $75$ .

$- 36$

$D_{z} = - 36$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $- 9$ for $D$ and $- 36$ for $D_{z}$ in the formula.

$z = \frac{- 36}{- 9}$

Step 6.5

Divide $- 36$ by $- 9$ .

$z = 4$

Step 7

List the solution to the system of equations.

$x = 1$

$y = - 1$

$z = 4$