Finite Math Examples

$x + y - z = 0$ , $x - y + z = 8$ , $x + 5 y - z = - 20$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 1 & - 1 \\ 1 & - 1 & 1 \\ 1 & 5 & - 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ 8 \\ - 20 \end{array}]$

Step 2

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

Tap for more steps...

Step 2.1

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

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

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.

Tap for more steps...

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Add the terms together.

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

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

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

Step 2.3.2

Simplify the determinant.

Tap for more steps...

Step 2.3.2.1

Simplify each term.

Tap for more steps...

Step 2.3.2.1.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.2.1.2

Multiply $- 5$ by $1$ .

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

Step 2.3.2.2

Subtract $5$ from $1$ .

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

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

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

Step 2.4.2

Simplify the determinant.

Tap for more steps...

Step 2.4.2.1

Simplify each term.

Tap for more steps...

Step 2.4.2.1.1

Multiply $- 1$ by $1$ .

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

Step 2.4.2.1.2

Multiply $- 1$ by $1$ .

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

Step 2.4.2.2

Subtract $1$ from $- 1$ .

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

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

Simplify the determinant.

Tap for more steps...

Step 2.5.2.1

Simplify each term.

Tap for more steps...

Step 2.5.2.1.1

Multiply $5$ by $1$ .

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

Step 2.5.2.1.2

Multiply $- 1$ by $- 1$ .

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

Step 2.5.2.2

Add $5$ and $1$ .

$1 \cdot - 4 - 1 \cdot - 2 - 1 \cdot 6$

Step 2.6

Simplify the determinant.

Tap for more steps...

Step 2.6.1

Simplify each term.

Tap for more steps...

Step 2.6.1.1

Multiply $- 4$ by $1$ .

$- 4 - 1 \cdot - 2 - 1 \cdot 6$

Step 2.6.1.2

Multiply $- 1$ by $- 2$ .

$- 4 + 2 - 1 \cdot 6$

Step 2.6.1.3

Multiply $- 1$ by $6$ .

$- 4 + 2 - 6$

Step 2.6.2

Add $- 4$ and $2$ .

$- 2 - 6$

Step 2.6.3

Subtract $6$ from $- 2$ .

$- 8$

$D = - 8$

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

Tap for more steps...

Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} 0 \\ 8 \\ - 20 \end{array}]$ .

$| \begin{array}{c} 0 & 1 & - 1 \\ 8 & - 1 & 1 \\ - 20 & 5 & - 1 \end{array} |$

Step 4.2

Find the determinant.

Tap for more steps...

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.

Tap for more steps...

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.

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

Step 4.2.1.4

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

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

Step 4.2.1.5

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

$| \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} |$

Step 4.2.1.6

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

$- 1 | \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} |$

Step 4.2.1.7

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

$| \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.1.8

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

$- 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.1.9

Add the terms together.

$0 | \begin{array}{c} - 1 & 1 \\ 5 & - 1 \end{array} | - 1 | \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} | - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.2

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

$0 - 1 | \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} | - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.3

Evaluate $| \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} |$ .

Tap for more steps...

Step 4.2.3.1

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

$0 - 1 (8 \cdot - 1 - (- 20 \cdot 1)) - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.3.2

Simplify the determinant.

Tap for more steps...

Step 4.2.3.2.1

Simplify each term.

Tap for more steps...

Step 4.2.3.2.1.1

Multiply $8$ by $- 1$ .

$0 - 1 (- 8 - (- 20 \cdot 1)) - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.3.2.1.2

Multiply $- (- 20 \cdot 1)$ .

Tap for more steps...

Step 4.2.3.2.1.2.1

Multiply $- 20$ by $1$ .

$0 - 1 (- 8 - - 20) - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.3.2.1.2.2

Multiply $- 1$ by $- 20$ .

$0 - 1 (- 8 + 20) - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.3.2.2

Add $- 8$ and $20$ .

$0 - 1 \cdot 12 - 1 | \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$

Step 4.2.4

Evaluate $| \begin{array}{c} 8 & - 1 \\ - 20 & 5 \end{array} |$ .

Tap for more steps...

Step 4.2.4.1

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

$0 - 1 \cdot 12 - 1 (8 \cdot 5 - (- 20 \cdot - 1))$

Step 4.2.4.2

Simplify the determinant.

Tap for more steps...

Step 4.2.4.2.1

Simplify each term.

Tap for more steps...

Step 4.2.4.2.1.1

Multiply $8$ by $5$ .

$0 - 1 \cdot 12 - 1 (40 - (- 20 \cdot - 1))$

Step 4.2.4.2.1.2

Multiply $- (- 20 \cdot - 1)$ .

Tap for more steps...

Step 4.2.4.2.1.2.1

Multiply $- 20$ by $- 1$ .

$0 - 1 \cdot 12 - 1 (40 - 1 \cdot 20)$

Step 4.2.4.2.1.2.2

Multiply $- 1$ by $20$ .

$0 - 1 \cdot 12 - 1 (40 - 20)$

Step 4.2.4.2.2

Subtract $20$ from $40$ .

$0 - 1 \cdot 12 - 1 \cdot 20$

Step 4.2.5

Simplify the determinant.

Tap for more steps...

Step 4.2.5.1

Simplify each term.

Tap for more steps...

Step 4.2.5.1.1

Multiply $- 1$ by $12$ .

$0 - 12 - 1 \cdot 20$

Step 4.2.5.1.2

Multiply $- 1$ by $20$ .

$0 - 12 - 20$

Step 4.2.5.2

Subtract $12$ from $0$ .

$- 12 - 20$

Step 4.2.5.3

Subtract $20$ from $- 12$ .

$- 32$

$D_{x} = - 32$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $- 8$ for $D$ and $- 32$ for $D_{x}$ in the formula.

$x = \frac{- 32}{- 8}$

Step 4.5

Divide $- 32$ by $- 8$ .

$x = 4$

Step 5

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

Tap for more steps...

Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} 0 \\ 8 \\ - 20 \end{array}]$ .

$| \begin{array}{c} 1 & 0 & - 1 \\ 1 & 8 & 1 \\ 1 & - 20 & - 1 \end{array} |$

Step 5.2

Find the determinant.

Tap for more steps...

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.

Tap for more steps...

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.

$| \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} |$

Step 5.2.1.4

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

$1 | \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} |$

Step 5.2.1.5

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

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

Step 5.2.1.6

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

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

Step 5.2.1.7

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

$| \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.1.8

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

$- 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.1.9

Add the terms together.

$1 | \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array} | - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.2

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

$1 | \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} | + 0 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.3

Evaluate $| \begin{array}{c} 8 & 1 \\ - 20 & - 1 \end{array} |$ .

Tap for more steps...

Step 5.2.3.1

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

$1 (8 \cdot - 1 - (- 20 \cdot 1)) + 0 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.3.2

Simplify the determinant.

Tap for more steps...

Step 5.2.3.2.1

Simplify each term.

Tap for more steps...

Step 5.2.3.2.1.1

Multiply $8$ by $- 1$ .

$1 (- 8 - (- 20 \cdot 1)) + 0 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.3.2.1.2

Multiply $- (- 20 \cdot 1)$ .

Tap for more steps...

Step 5.2.3.2.1.2.1

Multiply $- 20$ by $1$ .

$1 (- 8 - - 20) + 0 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.3.2.1.2.2

Multiply $- 1$ by $- 20$ .

$1 (- 8 + 20) + 0 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.3.2.2

Add $- 8$ and $20$ .

$1 \cdot 12 + 0 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$ .

Tap for more steps...

Step 5.2.4.1

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

$1 \cdot 12 + 0 - 1 (1 \cdot - 20 - 1 \cdot 8)$

Step 5.2.4.2

Simplify the determinant.

Tap for more steps...

Step 5.2.4.2.1

Simplify each term.

Tap for more steps...

Step 5.2.4.2.1.1

Multiply $- 20$ by $1$ .

$1 \cdot 12 + 0 - 1 (- 20 - 1 \cdot 8)$

Step 5.2.4.2.1.2

Multiply $- 1$ by $8$ .

$1 \cdot 12 + 0 - 1 (- 20 - 8)$

Step 5.2.4.2.2

Subtract $8$ from $- 20$ .

$1 \cdot 12 + 0 - 1 \cdot - 28$

Step 5.2.5

Simplify the determinant.

Tap for more steps...

Step 5.2.5.1

Simplify each term.

Tap for more steps...

Step 5.2.5.1.1

Multiply $12$ by $1$ .

$12 + 0 - 1 \cdot - 28$

Step 5.2.5.1.2

Multiply $- 1$ by $- 28$ .

$12 + 0 + 28$

Step 5.2.5.2

Add $12$ and $0$ .

$12 + 28$

Step 5.2.5.3

Add $12$ and $28$ .

$40$

$D_{y} = 40$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $- 8$ for $D$ and $40$ for $D_{y}$ in the formula.

$y = \frac{40}{- 8}$

Step 5.5

Divide $40$ by $- 8$ .

$y = - 5$

Step 6

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

Tap for more steps...

Step 6.1

Replace column $3$ of the coefficient matrix that corresponds to the $z$ -coefficients of the system with $[\begin{array}{c} 0 \\ 8 \\ - 20 \end{array}]$ .

$| \begin{array}{c} 1 & 1 & 0 \\ 1 & - 1 & 8 \\ 1 & 5 & - 20 \end{array} |$

Step 6.2

Find the determinant.

Tap for more steps...

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.

Tap for more steps...

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.

$| \begin{array}{c} - 1 & 8 \\ 5 & - 20 \end{array} |$

Step 6.2.1.4

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

$1 | \begin{array}{c} - 1 & 8 \\ 5 & - 20 \end{array} |$

Step 6.2.1.5

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

$| \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 6.2.1.6

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

$- 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$

Step 6.2.1.7

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

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

Step 6.2.1.8

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

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

Step 6.2.1.9

Add the terms together.

$1 | \begin{array}{c} - 1 & 8 \\ 5 & - 20 \end{array} | - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 1 & 5 \end{array} |$

Step 6.2.2

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

$1 | \begin{array}{c} - 1 & 8 \\ 5 & - 20 \end{array} | - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} | + 0$

Step 6.2.3

Evaluate $| \begin{array}{c} - 1 & 8 \\ 5 & - 20 \end{array} |$ .

Tap for more steps...

Step 6.2.3.1

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

$1 (- - 20 - 5 \cdot 8) - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} | + 0$

Step 6.2.3.2

Simplify the determinant.

Tap for more steps...

Step 6.2.3.2.1

Simplify each term.

Tap for more steps...

Step 6.2.3.2.1.1

Multiply $- 1$ by $- 20$ .

$1 (20 - 5 \cdot 8) - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} | + 0$

Step 6.2.3.2.1.2

Multiply $- 5$ by $8$ .

$1 (20 - 40) - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} | + 0$

Step 6.2.3.2.2

Subtract $40$ from $20$ .

$1 \cdot - 20 - 1 | \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} | + 0$

Step 6.2.4

Evaluate $| \begin{array}{c} 1 & 8 \\ 1 & - 20 \end{array} |$ .

Tap for more steps...

Step 6.2.4.1

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

$1 \cdot - 20 - 1 (1 \cdot - 20 - 1 \cdot 8) + 0$

Step 6.2.4.2

Simplify the determinant.

Tap for more steps...

Step 6.2.4.2.1

Simplify each term.

Tap for more steps...

Step 6.2.4.2.1.1

Multiply $- 20$ by $1$ .

$1 \cdot - 20 - 1 (- 20 - 1 \cdot 8) + 0$

Step 6.2.4.2.1.2

Multiply $- 1$ by $8$ .

$1 \cdot - 20 - 1 (- 20 - 8) + 0$

Step 6.2.4.2.2

Subtract $8$ from $- 20$ .

$1 \cdot - 20 - 1 \cdot - 28 + 0$

Step 6.2.5

Simplify the determinant.

Tap for more steps...

Step 6.2.5.1

Simplify each term.

Tap for more steps...

Step 6.2.5.1.1

Multiply $- 20$ by $1$ .

$- 20 - 1 \cdot - 28 + 0$

Step 6.2.5.1.2

Multiply $- 1$ by $- 28$ .

$- 20 + 28 + 0$

Step 6.2.5.2

Add $- 20$ and $28$ .

$8 + 0$

Step 6.2.5.3

Add $8$ and $0$ .

$8$

$D_{z} = 8$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $- 8$ for $D$ and $8$ for $D_{z}$ in the formula.

$z = \frac{8}{- 8}$

Step 6.5

Divide $8$ by $- 8$ .

$z = - 1$

Step 7

List the solution to the system of equations.

$x = 4$

$y = - 5$

$z = - 1$