Linear Algebra Examples

$[\begin{array}{c} 1 & 4 & 5 & 0 \\ 0 & 2 & 1 & 3 \\ 2 & 5 & 4 & 1 \\ 1 & 5 & 0 & 2 \end{array}]$

Step 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 1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{array} |$

Step 1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 1.4

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

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

Step 1.5

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

$| \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} |$

Step 1.6

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

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

Step 1.7

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

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

Step 1.8

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

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

Step 1.9

The minor for $a_{14}$ is the determinant with row $1$ and column $4$ deleted.

$| \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$

Step 1.10

Multiply element $a_{14}$ by its cofactor.

$0 | \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$

Step 1.11

Add the terms together.

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

Step 2

Multiply $0$ by $| \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$ .

Step 3

Evaluate $| \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |$ .

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Step 3.1

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 $2$ by its cofactor and add.

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Step 3.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 3.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 3.1.3

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

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

Step 3.1.4

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

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

Step 3.1.5

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

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

Step 3.1.6

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

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

Step 3.1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 3.1.8

Multiply element $a_{32}$ by its cofactor.

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

Step 3.1.9

Add the terms together.

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

Step 3.2

Multiply $0$ by $| \begin{array}{c} 2 & 3 \\ 5 & 1 \end{array} |$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the determinant.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $5$ by $2$ .

$1 (- 1 (10 - 5 \cdot 1) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.3.2.1.2

Multiply $- 5$ by $1$ .

$1 (- 1 (10 - 5) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.3.2.2

Subtract $5$ from $10$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify the determinant.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply $2$ by $2$ .

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

Step 3.4.2.1.2

Multiply $- 5$ by $3$ .

$1 (- 1 \cdot 5 + 4 (4 - 15) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.4.2.2

Subtract $15$ from $4$ .

$1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5

Simplify the determinant.

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Step 3.5.1

Simplify each term.

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Step 3.5.1.1

Multiply $- 1$ by $5$ .

$1 (- 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5.1.2

Multiply $4$ by $- 11$ .

$1 (- 5 - 44 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5.2

Subtract $44$ from $- 5$ .

$1 (- 49 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5.3

Add $- 49$ and $0$ .

$1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4

Evaluate $| \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} |$ .

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Step 4.1

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 4.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 4.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 4.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 4.1.4

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

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

Step 4.1.5

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

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

Step 4.1.6

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

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

Step 4.1.7

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

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

Step 4.1.8

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

$3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |$

Step 4.1.9

Add the terms together.

$1 \cdot - 49 - 4 (0 | \begin{array}{c} 4 & 1 \\ 0 & 2 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.2

Multiply $0$ by $| \begin{array}{c} 4 & 1 \\ 0 & 2 \end{array} |$ .

$1 \cdot - 49 - 4 (0 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3

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

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Step 4.3.1

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

$1 \cdot - 49 - 4 (0 - 1 (2 \cdot 2 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3.2

Simplify the determinant.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Multiply $2$ by $2$ .

$1 \cdot - 49 - 4 (0 - 1 (4 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3.2.1.2

Multiply $- 1$ by $1$ .

$1 \cdot - 49 - 4 (0 - 1 (4 - 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3.2.2

Subtract $1$ from $4$ .

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4

Evaluate $| \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |$ .

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Step 4.4.1

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

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (2 \cdot 0 - 1 \cdot 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4.2

Simplify the determinant.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

Multiply $2$ by $0$ .

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 1 \cdot 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4.2.1.2

Multiply $- 1$ by $4$ .

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4.2.2

Subtract $4$ from $0$ .

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5

Simplify the determinant.

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Step 4.5.1

Simplify each term.

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Step 4.5.1.1

Multiply $- 1$ by $3$ .

$1 \cdot - 49 - 4 (0 - 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5.1.2

Multiply $3$ by $- 4$ .

$1 \cdot - 49 - 4 (0 - 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5.2

Subtract $3$ from $0$ .

$1 \cdot - 49 - 4 (- 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5.3

Subtract $12$ from $- 3$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 5

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

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Step 5.1

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 5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 5.1.4

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

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

Step 5.1.5

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

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

Step 5.1.6

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

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

Step 5.1.7

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

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

Step 5.1.8

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

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

Step 5.1.9

Add the terms together.

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.2

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

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3

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

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Step 5.3.1

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

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (2 \cdot 2 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3.2

Simplify the determinant.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Multiply $2$ by $2$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3.2.1.2

Multiply $- 1$ by $1$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3.2.2

Subtract $1$ from $4$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.4

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

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Step 5.4.1

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

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (2 \cdot 5 - 1 \cdot 5)) + 0$

Step 5.4.2

Simplify the determinant.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Multiply $2$ by $5$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 1 \cdot 5)) + 0$

Step 5.4.2.1.2

Multiply $- 1$ by $5$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 5)) + 0$

Step 5.4.2.2

Subtract $5$ from $10$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

Multiply $- 2$ by $3$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 3 \cdot 5) + 0$

Step 5.5.1.2

Multiply $3$ by $5$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 15) + 0$

Step 5.5.2

Subtract $6$ from $0$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 (- 6 + 15) + 0$

Step 5.5.3

Add $- 6$ and $15$ .

$1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0$

Step 6

Simplify the determinant.

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Step 6.1

Simplify each term.

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Step 6.1.1

Multiply $- 49$ by $1$ .

$- 49 - 4 \cdot - 15 + 5 \cdot 9 + 0$

Step 6.1.2

Multiply $- 4$ by $- 15$ .

$- 49 + 60 + 5 \cdot 9 + 0$

Step 6.1.3

Multiply $5$ by $9$ .

$- 49 + 60 + 45 + 0$

Step 6.2

Add $- 49$ and $60$ .

$11 + 45 + 0$

Step 6.3

Add $11$ and $45$ .

$56 + 0$

Step 6.4

Add $56$ and $0$ .

$56$

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