Precalculus Examples

$2 w - x + 4 y - z = 1$ , $w + x - y + 2 z = 2$ , $3 w + 0 + 3 y + 1 z = 3$ , $3 w - 3 x + 9 y - 4 z = 0$

Step 1

Simplify.

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Add $3 w$ and $0$ .

$2 w - x + 4 y - z = 1, w + x - y + 2 z = 2, 3 w + 3 y + 1 z = 3, 3 w - 3 x + 9 y - 4 z = 0$

Multiply $z$ by $1$ .

$2 w - x + 4 y - z = 1, w + x - y + 2 z = 2, 3 w + 3 y + z = 3, 3 w - 3 x + 9 y - 4 z = 0$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 2 & - 1 & 4 & - 1 \\ 1 & 1 & - 1 & 2 \\ 3 & 0 & 3 & 1 \\ 3 & - 3 & 9 & - 4 \end{array}] [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array}]$

Step 3

Find the determinant of $[\begin{array}{c} 2 & - 1 & 4 & - 1 \\ 1 & 1 & - 1 & 2 \\ 3 & 0 & 3 & 1 \\ 3 & - 3 & 9 & - 4 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$D = 1 | \begin{array}{c} 1 & - 1 & 2 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify each term.

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Multiply $| \begin{array}{c} 1 & - 1 & 2 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} |$ by $1$ .

$D = | \begin{array}{c} 1 & - 1 & 2 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply every element in the $1 s t$ row by its cofactor and add.

$D = 1 | \begin{array}{c} 3 & 1 \\ 9 & - 4 \end{array} | + 1 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Evaluate $| \begin{array}{c} 3 & 1 \\ 9 & - 4 \end{array} |$ .

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

$D = 1 (3 \cdot - 4 - 9 \cdot 1) + 1 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 4$ .

$D = 1 (- 12 - 9 \cdot 1) + 1 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 9$ by $1$ .

$D = 1 (- 12 - 9) + 1 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $9$ from $- 12$ .

$D = 1 \cdot - 21 + 1 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

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

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

$D = 1 \cdot - 21 + 1 (3 \cdot - 4 - 3 \cdot 1) + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 4$ .

$D = 1 \cdot - 21 + 1 (- 12 - 3 \cdot 1) + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 3$ by $1$ .

$D = 1 \cdot - 21 + 1 (- 12 - 3) + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $3$ from $- 12$ .

$D = 1 \cdot - 21 + 1 \cdot - 15 + 2 | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Evaluate $| \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} |$ .

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

$D = 1 \cdot - 21 + 1 \cdot - 15 + 2 (3 \cdot 9 - 3 \cdot 3) + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $9$ .

$D = 1 \cdot - 21 + 1 \cdot - 15 + 2 (27 - 3 \cdot 3) + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 3$ by $3$ .

$D = 1 \cdot - 21 + 1 \cdot - 15 + 2 (27 - 9) + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $9$ from $27$ .

$D = 1 \cdot - 21 + 1 \cdot - 15 + 2 \cdot 18 + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 21$ by $1$ .

$D = - 21 + 1 \cdot - 15 + 2 \cdot 18 + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 15$ by $1$ .

$D = - 21 - 15 + 2 \cdot 18 + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $2$ by $18$ .

$D = - 21 - 15 + 36 + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $15$ from $- 21$ .

$D = - 36 + 36 + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Add $- 36$ and $36$ .

$D = 0 + 1 | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $| \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} |$ by $1$ .

$D = 0 + | \begin{array}{c} 2 & 4 & - 1 \\ 3 & 3 & 1 \\ 3 & 9 & - 4 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply every element in the $1 s t$ row by its cofactor and add.

$D = 0 + 2 | \begin{array}{c} 3 & 1 \\ 9 & - 4 \end{array} | - 4 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Evaluate $| \begin{array}{c} 3 & 1 \\ 9 & - 4 \end{array} |$ .

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

$D = 0 + 2 (3 \cdot - 4 - 9 \cdot 1) - 4 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 4$ .

$D = 0 + 2 (- 12 - 9 \cdot 1) - 4 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 9$ by $1$ .

$D = 0 + 2 (- 12 - 9) - 4 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $9$ from $- 12$ .

$D = 0 + 2 \cdot - 21 - 4 | \begin{array}{c} 3 & 1 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

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

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

$D = 0 + 2 \cdot - 21 - 4 (3 \cdot - 4 - 3 \cdot 1) - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 4$ .

$D = 0 + 2 \cdot - 21 - 4 (- 12 - 3 \cdot 1) - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 3$ by $1$ .

$D = 0 + 2 \cdot - 21 - 4 (- 12 - 3) - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $3$ from $- 12$ .

$D = 0 + 2 \cdot - 21 - 4 \cdot - 15 - | \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} | + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Evaluate $| \begin{array}{c} 3 & 3 \\ 3 & 9 \end{array} |$ .

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

$D = 0 + 2 \cdot - 21 - 4 \cdot - 15 - (3 \cdot 9 - 3 \cdot 3) + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $9$ .

$D = 0 + 2 \cdot - 21 - 4 \cdot - 15 - (27 - 3 \cdot 3) + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 3$ by $3$ .

$D = 0 + 2 \cdot - 21 - 4 \cdot - 15 - (27 - 9) + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $9$ from $27$ .

$D = 0 + 2 \cdot - 21 - 4 \cdot - 15 - 1 \cdot 18 + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $- 21$ .

$D = 0 - 42 - 4 \cdot - 15 - 1 \cdot 18 + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 4$ by $- 15$ .

$D = 0 - 42 + 60 - 1 \cdot 18 + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 1$ by $18$ .

$D = 0 - 42 + 60 - 18 + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Add $- 42$ and $60$ .

$D = 0 + 18 - 18 + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $18$ from $18$ .

$D = 0 + 0 + 0 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 9 & - 4 \end{array} | - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply every element in the $1 s t$ row by its cofactor and add.

$D = 0 + 0 + 0 (2 | \begin{array}{c} - 1 & 2 \\ 9 & - 4 \end{array} | - 4 | \begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Evaluate $| \begin{array}{c} - 1 & 2 \\ 9 & - 4 \end{array} |$ .

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

$D = 0 + 0 + 0 (2 (4 - 9 \cdot 2) - 4 | \begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $- 4$ .

Multiply $- 9$ by $2$ .

$D = 0 + 0 + 0 (2 (4 - 18) - 4 | \begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $18$ from $4$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 | \begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array} | - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

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

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

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 (1 \cdot - 4 - 3 \cdot 2) - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 4$ by $1$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 (- 4 - 3 \cdot 2) - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 3$ by $2$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 (- 4 - 6) - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $6$ from $- 4$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 \cdot - 10 - | \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Evaluate $| \begin{array}{c} 1 & - 1 \\ 3 & 9 \end{array} |$ .

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

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 \cdot - 10 - (1 \cdot 9 - 3 \cdot - 1)) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $9$ by $1$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 \cdot - 10 - (9 - 3 \cdot - 1)) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 3$ by $- 1$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 \cdot - 10 - (9 + 3)) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Add $9$ and $3$ .

$D = 0 + 0 + 0 (2 \cdot - 14 - 4 \cdot - 10 - 1 \cdot 12) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $- 14$ .

$D = 0 + 0 + 0 (- 28 - 4 \cdot - 10 - 1 \cdot 12) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 4$ by $- 10$ .

$D = 0 + 0 + 0 (- 28 + 40 - 1 \cdot 12) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $- 1$ by $12$ .

$D = 0 + 0 + 0 (- 28 + 40 - 12) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Add $- 28$ and $40$ .

$D = 0 + 0 + 0 (12 - 12) - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Subtract $12$ from $12$ .

$D = 0 + 0 + 0 \cdot 0 - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply $0$ by $0$ .

$D = 0 + 0 + 0 - 3 | \begin{array}{c} 2 & 4 & - 1 \\ 1 & - 1 & 2 \\ 3 & 3 & 1 \end{array} |$

Multiply every element in the $1 s t$ row by its cofactor and add.

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

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

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $1$ .

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

Multiply $- 3$ by $2$ .

$D = 0 + 0 + 0 - 3 (2 (- 1 - 6) - 4 | \begin{array}{c} 1 & 2 \\ 3 & 1 \end{array} | - | \begin{array}{c} 1 & - 1 \\ 3 & 3 \end{array} |)$

Subtract $6$ from $- 1$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 | \begin{array}{c} 1 & 2 \\ 3 & 1 \end{array} | - | \begin{array}{c} 1 & - 1 \\ 3 & 3 \end{array} |)$

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

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

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 (1 \cdot 1 - 3 \cdot 2) - | \begin{array}{c} 1 & - 1 \\ 3 & 3 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 (1 - 3 \cdot 2) - | \begin{array}{c} 1 & - 1 \\ 3 & 3 \end{array} |)$

Multiply $- 3$ by $2$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 (1 - 6) - | \begin{array}{c} 1 & - 1 \\ 3 & 3 \end{array} |)$

Subtract $6$ from $1$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 \cdot - 5 - | \begin{array}{c} 1 & - 1 \\ 3 & 3 \end{array} |)$

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

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

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 \cdot - 5 - (1 \cdot 3 - 3 \cdot - 1))$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $1$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 \cdot - 5 - (3 - 3 \cdot - 1))$

Multiply $- 3$ by $- 1$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 \cdot - 5 - (3 + 3))$

Add $3$ and $3$ .

$D = 0 + 0 + 0 - 3 (2 \cdot - 7 - 4 \cdot - 5 - 1 \cdot 6)$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $- 7$ .

$D = 0 + 0 + 0 - 3 (- 14 - 4 \cdot - 5 - 1 \cdot 6)$

Multiply $- 4$ by $- 5$ .

$D = 0 + 0 + 0 - 3 (- 14 + 20 - 1 \cdot 6)$

Multiply $- 1$ by $6$ .

$D = 0 + 0 + 0 - 3 (- 14 + 20 - 6)$

Add $- 14$ and $20$ .

$D = 0 + 0 + 0 - 3 (6 - 6)$

Subtract $6$ from $6$ .

$D = 0 + 0 + 0 - 3 \cdot 0$

Multiply $- 3$ by $0$ .

$D = 0 + 0 + 0 + 0$

Add $0$ and $0$ .

$D = 0 + 0 + 0$

Add $0$ and $0$ .

$D = 0 + 0$

Add $0$ and $0$ .

$D = 0$

Step 4

Cannot use Cramer's Rule because the determinant is $0$ .

Cannot solve using Cramer's Rule

Step 5

Choose two equations and eliminate one variable. In this case, eliminate $z$ .

$2 w - x + 4 y - z = 1$

$3 w + 3 y + z = 3$

Step 6

Eliminate $z$ from the system.

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Add the two equations together to eliminate $z$ from the system.

	$2$	$w$	$-$	$x$	$+$	$4$	$y$	$-$	$z$	$=$	$1$
$+$	$3$	$w$			$+$	$3$	$y$	$+$	$z$	$=$	$3$
	$5$	$w$	$-$	$x$	$+$	$7$	$y$			$=$	$4$

The resultant equation has $z$ eliminated.

$5 w - x + 7 y = 4$

Step 7

Choose another two equations and eliminate $z$ .

$w + x - y + 2 z = 2$

$3 w + 3 y + z = 3$

Step 8

Eliminate $z$ from the system.

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Multiply each equation by the value that makes the coefficients of $z$ opposite.

$w + x - y + 2 z = 2$

$(- 2) \cdot (3 w + 3 y + z) = (- 2) (3)$

Simplify.

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Simplify the left side.

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Simplify $(- 2) \cdot (3 w + 3 y + z)$ .

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Apply the distributive property.

$w + x - y + 2 z = 2$

$- 2 (3 w) - 2 (3 y) - 2 z = (- 2) (3)$

Simplify.

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Multiply $3$ by $- 2$ .

$w + x - y + 2 z = 2$

$- 6 w - 2 (3 y) - 2 z = (- 2) (3)$

Multiply $3$ by $- 2$ .

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = (- 2) (3)$

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = (- 2) (3)$

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = (- 2) (3)$

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = (- 2) (3)$

Simplify the right side.

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Multiply $- 2$ by $3$ .

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = - 6$

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = - 6$

$w + x - y + 2 z = 2$

$- 6 w - 6 y - 2 z = - 6$

Add the two equations together to eliminate $z$ from the system.

			$w$	$+$	$x$		$-$	$y$	$+$	$2$	$z$	$=$		$2$
$+$	$-$	$6$	$w$			$-$	$6$	$y$	$-$	$2$	$z$	$=$	$-$	$6$
	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$				$=$	$-$	$4$

The resultant equation has $z$ eliminated.

$- 5 w + x - 7 y = - 4$

Step 9

Take the resultant equations and eliminate another variable. In this case, eliminate $x$ .

$5 w - x + 7 y = 4$

$- 5 w + x - 7 y = - 4$

Step 10

Eliminate $x$ from the system.

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Add the two equations together to eliminate $x$ from the system.

		$5$	$w$	$-$	$x$	$+$	$7$	$y$	$=$		$4$
$+$	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$	$=$	$-$	$4$
								$0$	$=$		$0$

The resultant equation has $x$ eliminated.

$0 = 0$

Step 11

Because the resultant equation includes no variables and is true, the system of equations has an infinite number of solutions.

Infinite number of solutions

	$2$	$w$	$-$	$x$	$+$	$4$	$y$	$-$	$z$	$=$	$1$
$+$	$3$	$w$			$+$	$3$	$y$	$+$	$z$	$=$	$3$
	$5$	$w$	$-$	$x$	$+$	$7$	$y$			$=$	$4$

			$w$	$+$	$x$		$-$	$y$	$+$	$2$	$z$	$=$		$2$
$+$	$-$	$6$	$w$			$-$	$6$	$y$	$-$	$2$	$z$	$=$	$-$	$6$
	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$				$=$	$-$	$4$

		$5$	$w$	$-$	$x$	$+$	$7$	$y$	$=$		$4$
$+$	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$	$=$	$-$	$4$
								$0$	$=$		$0$

	$2$	$w$	$-$	$x$	$+$	$4$	$y$	$-$	$z$	$=$	$1$
$+$	$3$	$w$			$+$	$3$	$y$	$+$	$z$	$=$	$3$
	$5$	$w$	$-$	$x$	$+$	$7$	$y$			$=$	$4$

			$w$	$+$	$x$		$-$	$y$	$+$	$2$	$z$	$=$		$2$
$+$	$-$	$6$	$w$			$-$	$6$	$y$	$-$	$2$	$z$	$=$	$-$	$6$
	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$				$=$	$-$	$4$

		$5$	$w$	$-$	$x$	$+$	$7$	$y$	$=$		$4$
$+$	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$	$=$	$-$	$4$
								$0$	$=$		$0$

	$2$	$w$	$-$	$x$	$+$	$4$	$y$	$-$	$z$	$=$	$1$
$+$	$3$	$w$			$+$	$3$	$y$	$+$	$z$	$=$	$3$
	$5$	$w$	$-$	$x$	$+$	$7$	$y$			$=$	$4$

			$w$	$+$	$x$		$-$	$y$	$+$	$2$	$z$	$=$		$2$
$+$	$-$	$6$	$w$			$-$	$6$	$y$	$-$	$2$	$z$	$=$	$-$	$6$
	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$				$=$	$-$	$4$

		$5$	$w$	$-$	$x$	$+$	$7$	$y$	$=$		$4$
$+$	$-$	$5$	$w$	$+$	$x$	$-$	$7$	$y$	$=$	$-$	$4$
								$0$	$=$		$0$