Finite Math Examples

$x + 2 y - z = - 1$ , $2 x + y - z = 1$ , $x - 3 y - 5 z = 20$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 2 & - 1 \\ 2 & 1 & - 1 \\ 1 & - 3 & - 5 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 1 \\ 1 \\ 20 \end{array}]$

Step 2

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

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Step 2.1

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

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

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Add the terms together.

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

Step 2.3

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

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Step 2.3.1

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

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

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 $- 5$ by $1$ .

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

Step 2.3.2.1.2

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

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Step 2.3.2.1.2.1

Multiply $- 3$ by $- 1$ .

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

Step 2.3.2.1.2.2

Multiply $- 1$ by $3$ .

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

Step 2.3.2.2

Subtract $3$ from $- 5$ .

$1 \cdot - 8 - 2 | \begin{array}{c} 2 & - 1 \\ 1 & - 5 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 2.4

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

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Step 2.4.1

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

$1 \cdot - 8 - 2 (2 \cdot - 5 - 1 \cdot - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

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 $2$ by $- 5$ .

$1 \cdot - 8 - 2 (- 10 - 1 \cdot - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 2.4.2.1.2

Multiply $- 1$ by $- 1$ .

$1 \cdot - 8 - 2 (- 10 + 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 2.4.2.2

Add $- 10$ and $1$ .

$1 \cdot - 8 - 2 \cdot - 9 - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 2.5

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

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Step 2.5.1

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

$1 \cdot - 8 - 2 \cdot - 9 - 1 (2 \cdot - 3 - 1 \cdot 1)$

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 $- 3$ .

$1 \cdot - 8 - 2 \cdot - 9 - 1 (- 6 - 1 \cdot 1)$

Step 2.5.2.1.2

Multiply $- 1$ by $1$ .

$1 \cdot - 8 - 2 \cdot - 9 - 1 (- 6 - 1)$

Step 2.5.2.2

Subtract $1$ from $- 6$ .

$1 \cdot - 8 - 2 \cdot - 9 - 1 \cdot - 7$

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 $- 8$ by $1$ .

$- 8 - 2 \cdot - 9 - 1 \cdot - 7$

Step 2.6.1.2

Multiply $- 2$ by $- 9$ .

$- 8 + 18 - 1 \cdot - 7$

Step 2.6.1.3

Multiply $- 1$ by $- 7$ .

$- 8 + 18 + 7$

Step 2.6.2

Add $- 8$ and $18$ .

$10 + 7$

Step 2.6.3

Add $10$ and $7$ .

$17$

$D = 17$

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 $[\begin{array}{c} - 1 \\ 1 \\ 20 \end{array}]$ .

$| \begin{array}{c} - 1 & 2 & - 1 \\ 1 & 1 & - 1 \\ 20 & - 3 & - 5 \end{array} |$

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.

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

Step 4.2.1.4

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

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

Step 4.2.1.5

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

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

Step 4.2.1.6

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

$- 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} |$

Step 4.2.1.7

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

$| \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.1.8

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

$- 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.1.9

Add the terms together.

$- 1 | \begin{array}{c} 1 & - 1 \\ - 3 & - 5 \end{array} | - 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.2

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

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Step 4.2.2.1

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

$- 1 (1 \cdot - 5 - (- 3 \cdot - 1)) - 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

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 $1$ .

$- 1 (- 5 - (- 3 \cdot - 1)) - 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.2.2.1.2

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

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Step 4.2.2.2.1.2.1

Multiply $- 3$ by $- 1$ .

$- 1 (- 5 - 1 \cdot 3) - 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.2.2.1.2.2

Multiply $- 1$ by $3$ .

$- 1 (- 5 - 3) - 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.2.2.2

Subtract $3$ from $- 5$ .

$- 1 \cdot - 8 - 2 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.3

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

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Step 4.2.3.1

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

$- 1 \cdot - 8 - 2 (1 \cdot - 5 - 20 \cdot - 1) - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

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 $- 5$ by $1$ .

$- 1 \cdot - 8 - 2 (- 5 - 20 \cdot - 1) - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.3.2.1.2

Multiply $- 20$ by $- 1$ .

$- 1 \cdot - 8 - 2 (- 5 + 20) - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.3.2.2

Add $- 5$ and $20$ .

$- 1 \cdot - 8 - 2 \cdot 15 - 1 | \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$

Step 4.2.4

Evaluate $| \begin{array}{c} 1 & 1 \\ 20 & - 3 \end{array} |$ .

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Step 4.2.4.1

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

$- 1 \cdot - 8 - 2 \cdot 15 - 1 (1 \cdot - 3 - 20 \cdot 1)$

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 $- 3$ by $1$ .

$- 1 \cdot - 8 - 2 \cdot 15 - 1 (- 3 - 20 \cdot 1)$

Step 4.2.4.2.1.2

Multiply $- 20$ by $1$ .

$- 1 \cdot - 8 - 2 \cdot 15 - 1 (- 3 - 20)$

Step 4.2.4.2.2

Subtract $20$ from $- 3$ .

$- 1 \cdot - 8 - 2 \cdot 15 - 1 \cdot - 23$

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 $- 1$ by $- 8$ .

$8 - 2 \cdot 15 - 1 \cdot - 23$

Step 4.2.5.1.2

Multiply $- 2$ by $15$ .

$8 - 30 - 1 \cdot - 23$

Step 4.2.5.1.3

Multiply $- 1$ by $- 23$ .

$8 - 30 + 23$

Step 4.2.5.2

Subtract $30$ from $8$ .

$- 22 + 23$

Step 4.2.5.3

Add $- 22$ and $23$ .

$1$

$D_{x} = 1$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $17$ for $D$ and $1$ for $D_{x}$ in the formula.

$x = \frac{1}{17}$

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 $[\begin{array}{c} - 1 \\ 1 \\ 20 \end{array}]$ .

$| \begin{array}{c} 1 & - 1 & - 1 \\ 2 & 1 & - 1 \\ 1 & 20 & - 5 \end{array} |$

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.

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

Step 5.2.1.4

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

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

Step 5.2.1.5

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

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

Step 5.2.1.6

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

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

Step 5.2.1.7

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

$| \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.1.8

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

$- 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.1.9

Add the terms together.

$1 | \begin{array}{c} 1 & - 1 \\ 20 & - 5 \end{array} | + 1 | \begin{array}{c} 2 & - 1 \\ 1 & - 5 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.2

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

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Step 5.2.2.1

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

$1 (1 \cdot - 5 - 20 \cdot - 1) + 1 | \begin{array}{c} 2 & - 1 \\ 1 & - 5 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

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 $- 5$ by $1$ .

$1 (- 5 - 20 \cdot - 1) + 1 | \begin{array}{c} 2 & - 1 \\ 1 & - 5 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.2.2.1.2

Multiply $- 20$ by $- 1$ .

$1 (- 5 + 20) + 1 | \begin{array}{c} 2 & - 1 \\ 1 & - 5 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.2.2.2

Add $- 5$ and $20$ .

$1 \cdot 15 + 1 | \begin{array}{c} 2 & - 1 \\ 1 & - 5 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.3

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

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Step 5.2.3.1

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

$1 \cdot 15 + 1 (2 \cdot - 5 - 1 \cdot - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

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 $- 5$ .

$1 \cdot 15 + 1 (- 10 - 1 \cdot - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.3.2.1.2

Multiply $- 1$ by $- 1$ .

$1 \cdot 15 + 1 (- 10 + 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.3.2.2

Add $- 10$ and $1$ .

$1 \cdot 15 + 1 \cdot - 9 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$ .

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Step 5.2.4.1

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

$1 \cdot 15 + 1 \cdot - 9 - 1 (2 \cdot 20 - 1 \cdot 1)$

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 $2$ by $20$ .

$1 \cdot 15 + 1 \cdot - 9 - 1 (40 - 1 \cdot 1)$

Step 5.2.4.2.1.2

Multiply $- 1$ by $1$ .

$1 \cdot 15 + 1 \cdot - 9 - 1 (40 - 1)$

Step 5.2.4.2.2

Subtract $1$ from $40$ .

$1 \cdot 15 + 1 \cdot - 9 - 1 \cdot 39$

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 $15$ by $1$ .

$15 + 1 \cdot - 9 - 1 \cdot 39$

Step 5.2.5.1.2

Multiply $- 9$ by $1$ .

$15 - 9 - 1 \cdot 39$

Step 5.2.5.1.3

Multiply $- 1$ by $39$ .

$15 - 9 - 39$

Step 5.2.5.2

Subtract $9$ from $15$ .

$6 - 39$

Step 5.2.5.3

Subtract $39$ from $6$ .

$- 33$

$D_{y} = - 33$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $17$ for $D$ and $- 33$ for $D_{y}$ in the formula.

$y = \frac{- 33}{17}$

Step 5.5

Move the negative in front of the fraction.

$y = - \frac{33}{17}$

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 $[\begin{array}{c} - 1 \\ 1 \\ 20 \end{array}]$ .

$| \begin{array}{c} 1 & 2 & - 1 \\ 2 & 1 & 1 \\ 1 & - 3 & 20 \end{array} |$

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.

$| \begin{array}{c} 1 & 1 \\ - 3 & 20 \end{array} |$

Step 6.2.1.4

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

$1 | \begin{array}{c} 1 & 1 \\ - 3 & 20 \end{array} |$

Step 6.2.1.5

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

$| \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 6.2.1.6

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

$- 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$

Step 6.2.1.7

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

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

Step 6.2.1.8

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

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

Step 6.2.1.9

Add the terms together.

$1 | \begin{array}{c} 1 & 1 \\ - 3 & 20 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.2

Evaluate $| \begin{array}{c} 1 & 1 \\ - 3 & 20 \end{array} |$ .

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Step 6.2.2.1

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

$1 (1 \cdot 20 - (- 3 \cdot 1)) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

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 $20$ by $1$ .

$1 (20 - (- 3 \cdot 1)) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.2.2.1.2

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

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Step 6.2.2.2.1.2.1

Multiply $- 3$ by $1$ .

$1 (20 - - 3) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.2.2.1.2.2

Multiply $- 1$ by $- 3$ .

$1 (20 + 3) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.2.2.2

Add $20$ and $3$ .

$1 \cdot 23 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.3

Evaluate $| \begin{array}{c} 2 & 1 \\ 1 & 20 \end{array} |$ .

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Step 6.2.3.1

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

$1 \cdot 23 - 2 (2 \cdot 20 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

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 $20$ .

$1 \cdot 23 - 2 (40 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.3.2.1.2

Multiply $- 1$ by $1$ .

$1 \cdot 23 - 2 (40 - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.3.2.2

Subtract $1$ from $40$ .

$1 \cdot 23 - 2 \cdot 39 - 1 | \begin{array}{c} 2 & 1 \\ 1 & - 3 \end{array} |$

Step 6.2.4

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

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Step 6.2.4.1

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

$1 \cdot 23 - 2 \cdot 39 - 1 (2 \cdot - 3 - 1 \cdot 1)$

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 $- 3$ .

$1 \cdot 23 - 2 \cdot 39 - 1 (- 6 - 1 \cdot 1)$

Step 6.2.4.2.1.2

Multiply $- 1$ by $1$ .

$1 \cdot 23 - 2 \cdot 39 - 1 (- 6 - 1)$

Step 6.2.4.2.2

Subtract $1$ from $- 6$ .

$1 \cdot 23 - 2 \cdot 39 - 1 \cdot - 7$

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 $23$ by $1$ .

$23 - 2 \cdot 39 - 1 \cdot - 7$

Step 6.2.5.1.2

Multiply $- 2$ by $39$ .

$23 - 78 - 1 \cdot - 7$

Step 6.2.5.1.3

Multiply $- 1$ by $- 7$ .

$23 - 78 + 7$

Step 6.2.5.2

Subtract $78$ from $23$ .

$- 55 + 7$

Step 6.2.5.3

Add $- 55$ and $7$ .

$- 48$

$D_{z} = - 48$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $17$ for $D$ and $- 48$ for $D_{z}$ in the formula.

$z = \frac{- 48}{17}$

Step 6.5

Move the negative in front of the fraction.

$z = - \frac{48}{17}$

Step 7

List the solution to the system of equations.

$x = \frac{1}{17}$

$y = - \frac{33}{17}$

$z = - \frac{48}{17}$