Precalculus Examples

$2 x + y + z = 5$ , $4 x - y - 3 z = 1$ , $8 x + y - z = 5$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 2 & 1 & 1 \\ 4 & - 1 & - 3 \\ 8 & 1 & - 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 5 \\ 1 \\ 5 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 2 & 1 & 1 \\ 4 & - 1 & - 3 \\ 8 & 1 & - 1 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 2 & 1 & 1 \\ 4 & - 1 & - 3 \\ 8 & 1 & - 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} 2 & 1 & 1 \\ 4 & - 1 & - 3 \\ 8 & 1 & - 1 \end{array} |$

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

$| \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} |$

Step 2.2.6

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

$- 1 | \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} |$

Step 2.2.7

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

$| \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.2.8

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

$1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.2.9

Add the terms together.

$2 | \begin{array}{c} - 1 & - 3 \\ 1 & - 1 \end{array} | - 1 | \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.3

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

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Step 2.3.1

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

$2 (- - 1 - 1 \cdot - 3) - 1 | \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

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 $- 1$ .

$2 (1 - 1 \cdot - 3) - 1 | \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.3.2.1.2

Multiply $- 1$ by $- 3$ .

$2 (1 + 3) - 1 | \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.3.2.2

Add $1$ and $3$ .

$2 \cdot 4 - 1 | \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} | + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.4

Evaluate $| \begin{array}{c} 4 & - 3 \\ 8 & - 1 \end{array} |$ .

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Step 2.4.1

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

$2 \cdot 4 - 1 (4 \cdot - 1 - 8 \cdot - 3) + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

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 $4$ by $- 1$ .

$2 \cdot 4 - 1 (- 4 - 8 \cdot - 3) + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.4.2.1.2

Multiply $- 8$ by $- 3$ .

$2 \cdot 4 - 1 (- 4 + 24) + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.4.2.2

Add $- 4$ and $24$ .

$2 \cdot 4 - 1 \cdot 20 + 1 | \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$

Step 2.5

Evaluate $| \begin{array}{c} 4 & - 1 \\ 8 & 1 \end{array} |$ .

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Step 2.5.1

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

$2 \cdot 4 - 1 \cdot 20 + 1 (4 \cdot 1 - 8 \cdot - 1)$

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 $4$ by $1$ .

$2 \cdot 4 - 1 \cdot 20 + 1 (4 - 8 \cdot - 1)$

Step 2.5.2.1.2

Multiply $- 8$ by $- 1$ .

$2 \cdot 4 - 1 \cdot 20 + 1 (4 + 8)$

Step 2.5.2.2

Add $4$ and $8$ .

$2 \cdot 4 - 1 \cdot 20 + 1 \cdot 12$

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 $2$ by $4$ .

$8 - 1 \cdot 20 + 1 \cdot 12$

Step 2.6.1.2

Multiply $- 1$ by $20$ .

$8 - 20 + 1 \cdot 12$

Step 2.6.1.3

Multiply $12$ by $1$ .

$8 - 20 + 12$

Step 2.6.2

Subtract $20$ from $8$ .

$- 12 + 12$

Step 2.6.3

Add $- 12$ and $12$ .

$0$

$D = 0$

Step 3

Since the determinant is $0$ , the system cannot be solved using Cramer's Rule.