Precalculus Examples

$5 x + 2 y - z = 1$ , $x - 2 y + 2 z = 0$ , $2 x - 3 y + z = - 2$

Step 1

Represent the system of equations in matrix format.

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

Step 2

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

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Step 2.1

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

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

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Add the terms together.

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

Step 2.3

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

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Step 2.3.1

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

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

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 $1$ .

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

Step 2.3.2.1.2

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

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Step 2.3.2.1.2.1

Multiply $- 3$ by $2$ .

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

Step 2.3.2.1.2.2

Multiply $- 1$ by $- 6$ .

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

Step 2.3.2.2

Add $- 2$ and $6$ .

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

Step 2.4

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

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Step 2.4.1

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

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

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 $1$ by $1$ .

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

Step 2.4.2.1.2

Multiply $- 2$ by $2$ .

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

Step 2.4.2.2

Subtract $4$ from $1$ .

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

Step 2.5

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

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Step 2.5.1

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

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

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 $- 3$ by $1$ .

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

Step 2.5.2.1.2

Multiply $- 2$ by $- 2$ .

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

Step 2.5.2.2

Add $- 3$ and $4$ .

$5 \cdot 4 - 2 \cdot - 3 - 1 \cdot 1$

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 $5$ by $4$ .

$20 - 2 \cdot - 3 - 1 \cdot 1$

Step 2.6.1.2

Multiply $- 2$ by $- 3$ .

$20 + 6 - 1 \cdot 1$

Step 2.6.1.3

Multiply $- 1$ by $1$ .

$20 + 6 - 1$

Step 2.6.2

Add $20$ and $6$ .

$26 - 1$

Step 2.6.3

Subtract $1$ from $26$ .

$25$

$D = 25$

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 $[\begin{array}{c} 1 \\ 0 \\ - 2 \end{array}]$ .

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

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 column $1$ 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.

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

Step 4.2.1.4

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

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

Step 4.2.1.5

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

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

Step 4.2.1.6

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

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

Step 4.2.1.7

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

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

Step 4.2.1.8

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

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

Step 4.2.1.9

Add the terms together.

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

Step 4.2.2

Multiply $0$ by $| \begin{array}{c} 2 & - 1 \\ - 3 & 1 \end{array} |$ .

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

Step 4.2.3

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

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Step 4.2.3.1

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

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

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 $1$ .

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

Step 4.2.3.2.1.2

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

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Step 4.2.3.2.1.2.1

Multiply $- 3$ by $2$ .

$1 (- 2 - - 6) + 0 - 2 | \begin{array}{c} 2 & - 1 \\ - 2 & 2 \end{array} |$

Step 4.2.3.2.1.2.2

Multiply $- 1$ by $- 6$ .

$1 (- 2 + 6) + 0 - 2 | \begin{array}{c} 2 & - 1 \\ - 2 & 2 \end{array} |$

Step 4.2.3.2.2

Add $- 2$ and $6$ .

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

Step 4.2.4

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

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Step 4.2.4.1

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

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

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 $2$ .

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

Step 4.2.4.2.1.2

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

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Step 4.2.4.2.1.2.1

Multiply $- 2$ by $- 1$ .

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

Step 4.2.4.2.1.2.2

Multiply $- 1$ by $2$ .

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

Step 4.2.4.2.2

Subtract $2$ from $4$ .

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

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 $4$ by $1$ .

$4 + 0 - 2 \cdot 2$

Step 4.2.5.1.2

Multiply $- 2$ by $2$ .

$4 + 0 - 4$

Step 4.2.5.2

Add $4$ and $0$ .

$4 - 4$

Step 4.2.5.3

Subtract $4$ from $4$ .

$0$

$D_{x} = 0$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $25$ for $D$ and $0$ for $D_{x}$ in the formula.

$x = \frac{0}{25}$

Step 4.5

Divide $0$ by $25$ .

$x = 0$

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 $[\begin{array}{c} 1 \\ 0 \\ - 2 \end{array}]$ .

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

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

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

Step 5.2.1.4

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

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

Step 5.2.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 5.2.1.6

Multiply element $a_{22}$ by its cofactor.

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

Step 5.2.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

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

Step 5.2.1.8

Multiply element $a_{23}$ by its cofactor.

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

Step 5.2.1.9

Add the terms together.

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

Step 5.2.2

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

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

Step 5.2.3

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

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Step 5.2.3.1

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

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

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 $1$ by $1$ .

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

Step 5.2.3.2.1.2

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

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Step 5.2.3.2.1.2.1

Multiply $- 2$ by $- 1$ .

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

Step 5.2.3.2.1.2.2

Multiply $- 1$ by $2$ .

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

Step 5.2.3.2.2

Subtract $2$ from $1$ .

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

Step 5.2.4

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

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Step 5.2.4.1

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

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

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 $- 2$ .

$- 1 \cdot - 1 + 0 - 2 (- 10 - 2 \cdot 1)$

Step 5.2.4.2.1.2

Multiply $- 2$ by $1$ .

$- 1 \cdot - 1 + 0 - 2 (- 10 - 2)$

Step 5.2.4.2.2

Subtract $2$ from $- 10$ .

$- 1 \cdot - 1 + 0 - 2 \cdot - 12$

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 $- 1$ by $- 1$ .

$1 + 0 - 2 \cdot - 12$

Step 5.2.5.1.2

Multiply $- 2$ by $- 12$ .

$1 + 0 + 24$

Step 5.2.5.2

Add $1$ and $0$ .

$1 + 24$

Step 5.2.5.3

Add $1$ and $24$ .

$25$

$D_{y} = 25$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $25$ for $D$ and $25$ for $D_{y}$ in the formula.

$y = \frac{25}{25}$

Step 5.5

Divide $25$ by $25$ .

$y = 1$

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 $[\begin{array}{c} 1 \\ 0 \\ - 2 \end{array}]$ .

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

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

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

Step 6.2.1.4

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

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

Step 6.2.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 6.2.1.6

Multiply element $a_{22}$ by its cofactor.

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

Step 6.2.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

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

Step 6.2.1.8

Multiply element $a_{23}$ by its cofactor.

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

Step 6.2.1.9

Add the terms together.

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

Step 6.2.2

Multiply $0$ by $| \begin{array}{c} 5 & 2 \\ 2 & - 3 \end{array} |$ .

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

Step 6.2.3

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

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Step 6.2.3.1

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

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

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 $2$ by $- 2$ .

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

Step 6.2.3.2.1.2

Multiply $- (- 3 \cdot 1)$ .

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Step 6.2.3.2.1.2.1

Multiply $- 3$ by $1$ .

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

Step 6.2.3.2.1.2.2

Multiply $- 1$ by $- 3$ .

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

Step 6.2.3.2.2

Add $- 4$ and $3$ .

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

Step 6.2.4

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

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Step 6.2.4.1

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

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

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 $5$ by $- 2$ .

$- 1 \cdot - 1 - 2 (- 10 - 2 \cdot 1) + 0$

Step 6.2.4.2.1.2

Multiply $- 2$ by $1$ .

$- 1 \cdot - 1 - 2 (- 10 - 2) + 0$

Step 6.2.4.2.2

Subtract $2$ from $- 10$ .

$- 1 \cdot - 1 - 2 \cdot - 12 + 0$

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 $- 1$ by $- 1$ .

$1 - 2 \cdot - 12 + 0$

Step 6.2.5.1.2

Multiply $- 2$ by $- 12$ .

$1 + 24 + 0$

Step 6.2.5.2

Add $1$ and $24$ .

$25 + 0$

Step 6.2.5.3

Add $25$ and $0$ .

$25$

$D_{z} = 25$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $25$ for $D$ and $25$ for $D_{z}$ in the formula.

$z = \frac{25}{25}$

Step 6.5

Divide $25$ by $25$ .

$z = 1$

Step 7

List the solution to the system of equations.

$x = 0$

$y = 1$

$z = 1$