Finite Math Examples

$x - y + 3 z = 0$ , $2 x + y + 2 z = - 2$ , $- 2 x - 2 y + z = 7$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & - 1 & 3 \\ 2 & 1 & 2 \\ - 2 & - 2 & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ - 2 \\ 7 \end{array}]$

Step 2

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

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Step 2.1

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

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

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Add the terms together.

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

Step 2.3

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

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Step 2.3.1

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

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

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 $1$ by $1$ .

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

Step 2.3.2.1.2

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

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Step 2.3.2.1.2.1

Multiply $- 2$ by $2$ .

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

Step 2.3.2.1.2.2

Multiply $- 1$ by $- 4$ .

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

Step 2.3.2.2

Add $1$ and $4$ .

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

Step 2.4

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

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Step 2.4.1

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

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

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 $1$ .

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

Step 2.4.2.1.2

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

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Step 2.4.2.1.2.1

Multiply $- 2$ by $2$ .

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

Step 2.4.2.1.2.2

Multiply $- 1$ by $- 4$ .

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

Step 2.4.2.2

Add $2$ and $4$ .

$1 \cdot 5 + 1 \cdot 6 + 3 | \begin{array}{c} 2 & 1 \\ - 2 & - 2 \end{array} |$

Step 2.5

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

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Step 2.5.1

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

$1 \cdot 5 + 1 \cdot 6 + 3 (2 \cdot - 2 - (- 2 \cdot 1))$

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 $- 2$ .

$1 \cdot 5 + 1 \cdot 6 + 3 (- 4 - (- 2 \cdot 1))$

Step 2.5.2.1.2

Multiply $- (- 2 \cdot 1)$ .

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Step 2.5.2.1.2.1

Multiply $- 2$ by $1$ .

$1 \cdot 5 + 1 \cdot 6 + 3 (- 4 - - 2)$

Step 2.5.2.1.2.2

Multiply $- 1$ by $- 2$ .

$1 \cdot 5 + 1 \cdot 6 + 3 (- 4 + 2)$

Step 2.5.2.2

Add $- 4$ and $2$ .

$1 \cdot 5 + 1 \cdot 6 + 3 \cdot - 2$

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 $5$ by $1$ .

$5 + 1 \cdot 6 + 3 \cdot - 2$

Step 2.6.1.2

Multiply $6$ by $1$ .

$5 + 6 + 3 \cdot - 2$

Step 2.6.1.3

Multiply $3$ by $- 2$ .

$5 + 6 - 6$

Step 2.6.2

Add $5$ and $6$ .

$11 - 6$

Step 2.6.3

Subtract $6$ from $11$ .

$5$

$D = 5$

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 $[\begin{array}{c} 0 \\ - 2 \\ 7 \end{array}]$ .

$| \begin{array}{c} 0 & - 1 & 3 \\ - 2 & 1 & 2 \\ 7 & - 2 & 1 \end{array} |$

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.

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

Step 4.2.1.4

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

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

Step 4.2.1.5

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

$| \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} |$

Step 4.2.1.6

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

$1 | \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} |$

Step 4.2.1.7

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

$| \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.1.8

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

$3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.1.9

Add the terms together.

$0 | \begin{array}{c} 1 & 2 \\ - 2 & 1 \end{array} | + 1 | \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.2

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

$0 + 1 | \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.3

Evaluate $| \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} |$ .

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Step 4.2.3.1

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

$0 + 1 (- 2 \cdot 1 - 7 \cdot 2) + 3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

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 $- 2$ by $1$ .

$0 + 1 (- 2 - 7 \cdot 2) + 3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.3.2.1.2

Multiply $- 7$ by $2$ .

$0 + 1 (- 2 - 14) + 3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.3.2.2

Subtract $14$ from $- 2$ .

$0 + 1 \cdot - 16 + 3 | \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$

Step 4.2.4

Evaluate $| \begin{array}{c} - 2 & 1 \\ 7 & - 2 \end{array} |$ .

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Step 4.2.4.1

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

$0 + 1 \cdot - 16 + 3 (- 2 \cdot - 2 - 7 \cdot 1)$

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 $- 2$ by $- 2$ .

$0 + 1 \cdot - 16 + 3 (4 - 7 \cdot 1)$

Step 4.2.4.2.1.2

Multiply $- 7$ by $1$ .

$0 + 1 \cdot - 16 + 3 (4 - 7)$

Step 4.2.4.2.2

Subtract $7$ from $4$ .

$0 + 1 \cdot - 16 + 3 \cdot - 3$

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 $- 16$ by $1$ .

$0 - 16 + 3 \cdot - 3$

Step 4.2.5.1.2

Multiply $3$ by $- 3$ .

$0 - 16 - 9$

Step 4.2.5.2

Subtract $16$ from $0$ .

$- 16 - 9$

Step 4.2.5.3

Subtract $9$ from $- 16$ .

$- 25$

$D_{x} = - 25$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $5$ for $D$ and $- 25$ for $D_{x}$ in the formula.

$x = \frac{- 25}{5}$

Step 4.5

Divide $- 25$ by $5$ .

$x = - 5$

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 $[\begin{array}{c} 0 \\ - 2 \\ 7 \end{array}]$ .

$| \begin{array}{c} 1 & 0 & 3 \\ 2 & - 2 & 2 \\ - 2 & 7 & 1 \end{array} |$

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.

$| \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} |$

Step 5.2.1.4

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

$1 | \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} |$

Step 5.2.1.5

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

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

Step 5.2.1.6

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

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

Step 5.2.1.7

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

$| \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.1.8

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

$3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.1.9

Add the terms together.

$1 | \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 2 \\ - 2 & 1 \end{array} | + 3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.2

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

$1 | \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} | + 0 + 3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.3

Evaluate $| \begin{array}{c} - 2 & 2 \\ 7 & 1 \end{array} |$ .

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Step 5.2.3.1

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

$1 (- 2 \cdot 1 - 7 \cdot 2) + 0 + 3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

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 $1$ .

$1 (- 2 - 7 \cdot 2) + 0 + 3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.3.2.1.2

Multiply $- 7$ by $2$ .

$1 (- 2 - 14) + 0 + 3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.3.2.2

Subtract $14$ from $- 2$ .

$1 \cdot - 16 + 0 + 3 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$ .

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Step 5.2.4.1

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

$1 \cdot - 16 + 0 + 3 (2 \cdot 7 - (- 2 \cdot - 2))$

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 $7$ .

$1 \cdot - 16 + 0 + 3 (14 - (- 2 \cdot - 2))$

Step 5.2.4.2.1.2

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

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Step 5.2.4.2.1.2.1

Multiply $- 2$ by $- 2$ .

$1 \cdot - 16 + 0 + 3 (14 - 1 \cdot 4)$

Step 5.2.4.2.1.2.2

Multiply $- 1$ by $4$ .

$1 \cdot - 16 + 0 + 3 (14 - 4)$

Step 5.2.4.2.2

Subtract $4$ from $14$ .

$1 \cdot - 16 + 0 + 3 \cdot 10$

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 $- 16$ by $1$ .

$- 16 + 0 + 3 \cdot 10$

Step 5.2.5.1.2

Multiply $3$ by $10$ .

$- 16 + 0 + 30$

Step 5.2.5.2

Add $- 16$ and $0$ .

$- 16 + 30$

Step 5.2.5.3

Add $- 16$ and $30$ .

$14$

$D_{y} = 14$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $5$ for $D$ and $14$ for $D_{y}$ in the formula.

$y = \frac{14}{5}$

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 $[\begin{array}{c} 0 \\ - 2 \\ 7 \end{array}]$ .

$| \begin{array}{c} 1 & - 1 & 0 \\ 2 & 1 & - 2 \\ - 2 & - 2 & 7 \end{array} |$

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.

$| \begin{array}{c} 1 & - 2 \\ - 2 & 7 \end{array} |$

Step 6.2.1.4

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

$1 | \begin{array}{c} 1 & - 2 \\ - 2 & 7 \end{array} |$

Step 6.2.1.5

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

$| \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 6.2.1.6

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

$1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$

Step 6.2.1.7

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

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

Step 6.2.1.8

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

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

Step 6.2.1.9

Add the terms together.

$1 | \begin{array}{c} 1 & - 2 \\ - 2 & 7 \end{array} | + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0 | \begin{array}{c} 2 & 1 \\ - 2 & - 2 \end{array} |$

Step 6.2.2

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

$1 | \begin{array}{c} 1 & - 2 \\ - 2 & 7 \end{array} | + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0$

Step 6.2.3

Evaluate $| \begin{array}{c} 1 & - 2 \\ - 2 & 7 \end{array} |$ .

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Step 6.2.3.1

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

$1 (1 \cdot 7 - (- 2 \cdot - 2)) + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0$

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 $7$ by $1$ .

$1 (7 - (- 2 \cdot - 2)) + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0$

Step 6.2.3.2.1.2

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

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Step 6.2.3.2.1.2.1

Multiply $- 2$ by $- 2$ .

$1 (7 - 1 \cdot 4) + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0$

Step 6.2.3.2.1.2.2

Multiply $- 1$ by $4$ .

$1 (7 - 4) + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0$

Step 6.2.3.2.2

Subtract $4$ from $7$ .

$1 \cdot 3 + 1 | \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} | + 0$

Step 6.2.4

Evaluate $| \begin{array}{c} 2 & - 2 \\ - 2 & 7 \end{array} |$ .

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Step 6.2.4.1

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

$1 \cdot 3 + 1 (2 \cdot 7 - (- 2 \cdot - 2)) + 0$

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 $7$ .

$1 \cdot 3 + 1 (14 - (- 2 \cdot - 2)) + 0$

Step 6.2.4.2.1.2

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

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Step 6.2.4.2.1.2.1

Multiply $- 2$ by $- 2$ .

$1 \cdot 3 + 1 (14 - 1 \cdot 4) + 0$

Step 6.2.4.2.1.2.2

Multiply $- 1$ by $4$ .

$1 \cdot 3 + 1 (14 - 4) + 0$

Step 6.2.4.2.2

Subtract $4$ from $14$ .

$1 \cdot 3 + 1 \cdot 10 + 0$

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 $3$ by $1$ .

$3 + 1 \cdot 10 + 0$

Step 6.2.5.1.2

Multiply $10$ by $1$ .

$3 + 10 + 0$

Step 6.2.5.2

Add $3$ and $10$ .

$13 + 0$

Step 6.2.5.3

Add $13$ and $0$ .

$13$

$D_{z} = 13$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $5$ for $D$ and $13$ for $D_{z}$ in the formula.

$z = \frac{13}{5}$

Step 7

List the solution to the system of equations.

$x = - 5$

$y = \frac{14}{5}$

$z = \frac{13}{5}$