Finite Math Examples

$5 x + 8 y - 6 z = 14$ , $3 x + 4 y - 2 z = 8$ , $x + 2 y - 2 z = 3$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 5 & 8 & - 6 \\ 3 & 4 & - 2 \\ 1 & 2 & - 2 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 14 \\ 8 \\ 3 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 5 & 8 & - 6 \\ 3 & 4 & - 2 \\ 1 & 2 & - 2 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 5 & 8 & - 6 \\ 3 & 4 & - 2 \\ 1 & 2 & - 2 \end{array}]$ in determinant notation.

$| \begin{array}{c} 5 & 8 & - 6 \\ 3 & 4 & - 2 \\ 1 & 2 & - 2 \end{array} |$

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

$- 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.2.9

Add the terms together.

$5 | \begin{array}{c} 4 & - 2 \\ 2 & - 2 \end{array} | - 8 | \begin{array}{c} 3 & - 2 \\ 1 & - 2 \end{array} | - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.3

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

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Step 2.3.1

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

$5 (4 \cdot - 2 - 2 \cdot - 2) - 8 | \begin{array}{c} 3 & - 2 \\ 1 & - 2 \end{array} | - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

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 $4$ by $- 2$ .

$5 (- 8 - 2 \cdot - 2) - 8 | \begin{array}{c} 3 & - 2 \\ 1 & - 2 \end{array} | - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.3.2.1.2

Multiply $- 2$ by $- 2$ .

$5 (- 8 + 4) - 8 | \begin{array}{c} 3 & - 2 \\ 1 & - 2 \end{array} | - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.3.2.2

Add $- 8$ and $4$ .

$5 \cdot - 4 - 8 | \begin{array}{c} 3 & - 2 \\ 1 & - 2 \end{array} | - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.4

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

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Step 2.4.1

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

$5 \cdot - 4 - 8 (3 \cdot - 2 - 1 \cdot - 2) - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

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 $3$ by $- 2$ .

$5 \cdot - 4 - 8 (- 6 - 1 \cdot - 2) - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.4.2.1.2

Multiply $- 1$ by $- 2$ .

$5 \cdot - 4 - 8 (- 6 + 2) - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.4.2.2

Add $- 6$ and $2$ .

$5 \cdot - 4 - 8 \cdot - 4 - 6 | \begin{array}{c} 3 & 4 \\ 1 & 2 \end{array} |$

Step 2.5

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

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Step 2.5.1

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

$5 \cdot - 4 - 8 \cdot - 4 - 6 (3 \cdot 2 - 1 \cdot 4)$

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 $2$ .

$5 \cdot - 4 - 8 \cdot - 4 - 6 (6 - 1 \cdot 4)$

Step 2.5.2.1.2

Multiply $- 1$ by $4$ .

$5 \cdot - 4 - 8 \cdot - 4 - 6 (6 - 4)$

Step 2.5.2.2

Subtract $4$ from $6$ .

$5 \cdot - 4 - 8 \cdot - 4 - 6 \cdot 2$

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 $- 4$ .

$- 20 - 8 \cdot - 4 - 6 \cdot 2$

Step 2.6.1.2

Multiply $- 8$ by $- 4$ .

$- 20 + 32 - 6 \cdot 2$

Step 2.6.1.3

Multiply $- 6$ by $2$ .

$- 20 + 32 - 12$

Step 2.6.2

Add $- 20$ and $32$ .

$12 - 12$

Step 2.6.3

Subtract $12$ from $12$ .

$0$

$D = 0$

Step 3

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