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

Set up the determinant by breaking it into smaller components.

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

Step 2

Find the determinant of $1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |$ .

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Set up the determinant by breaking it into smaller components.

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

Find the determinant of $- 1 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} |$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $2$ .

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

Multiply $- 5$ by $1$ .

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

Simplify the expression.

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Subtract $5$ from $10$ .

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

Multiply $- 1$ by $5$ .

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

Find the determinant of $4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} |$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $2$ .

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

Multiply $- 5$ by $3$ .

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

Simplify the expression.

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Subtract $15$ from $4$ .

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

Multiply $4$ by $- 11$ .

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

Since the matrix is multiplied by $0$ , the determinant is $0$ .

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

Subtract $44$ from $- 5$ .

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

Add $- 49$ and $0$ .

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

Step 3

Since the matrix is multiplied by $0$ , the determinant is $0$ .

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

Step 4

Find the determinant of $2 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 0 & 2 \end{array} |$ .

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Set up the determinant by breaking it into smaller components.

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

Find the determinant of $- 5 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} |$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $2$ .

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

Multiply $- 5$ by $3$ .

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

Simplify the expression.

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Subtract $15$ from $4$ .

$- 49 + 0 + 2 (- 5 \cdot - 11 + 1 | \begin{array}{c} 4 & 0 \\ 5 & 2 \end{array} | + 0 | \begin{array}{c} 4 & 0 \\ 2 & 3 \end{array} |) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Multiply $- 5$ by $- 11$ .

$- 49 + 0 + 2 (55 + 1 | \begin{array}{c} 4 & 0 \\ 5 & 2 \end{array} | + 0 | \begin{array}{c} 4 & 0 \\ 2 & 3 \end{array} |) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Find the determinant of $1 | \begin{array}{c} 4 & 0 \\ 5 & 2 \end{array} |$ .

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

$- 49 + 0 + 2 (55 + (4) (2) - 5 \cdot 0 + 0 | \begin{array}{c} 4 & 0 \\ 2 & 3 \end{array} |) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $4$ by $2$ .

$- 49 + 0 + 2 (55 + 8 - 5 \cdot 0 + 0 | \begin{array}{c} 4 & 0 \\ 2 & 3 \end{array} |) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Multiply $- 5$ by $0$ .

$- 49 + 0 + 2 (55 + 8 + 0 + 0 | \begin{array}{c} 4 & 0 \\ 2 & 3 \end{array} |) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Add $8$ and $0$ .

$- 49 + 0 + 2 (55 + 8 + 0 | \begin{array}{c} 4 & 0 \\ 2 & 3 \end{array} |) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$- 49 + 0 + 2 (55 + 8 + 0) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Add $55$ and $8$ .

$- 49 + 0 + 2 (63 + 0) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Add $63$ and $0$ .

$- 49 + 0 + 2 (63) - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Multiply $2$ by $63$ .

$- 49 + 0 + 126 - 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$

Step 5

Find the determinant of $- 1 | \begin{array}{c} 4 & 5 & 0 \\ 2 & 1 & 3 \\ 5 & 4 & 1 \end{array} |$ .

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Set up the determinant by breaking it into smaller components.

$- 49 + 0 + 126 - 1 (0 | \begin{array}{c} 2 & 1 \\ 5 & 4 \end{array} | - 3 | \begin{array}{c} 4 & 5 \\ 5 & 4 \end{array} | + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$- 49 + 0 + 126 - 1 (0 - 3 | \begin{array}{c} 4 & 5 \\ 5 & 4 \end{array} | + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Find the determinant of $- 3 | \begin{array}{c} 4 & 5 \\ 5 & 4 \end{array} |$ .

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

$- 49 + 0 + 126 - 1 (0 - 3 ((4) (4) - 5 \cdot 5) + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $4$ by $4$ .

$- 49 + 0 + 126 - 1 (0 - 3 (16 - 5 \cdot 5) + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Multiply $- 5$ by $5$ .

$- 49 + 0 + 126 - 1 (0 - 3 (16 - 25) + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Simplify the expression.

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Subtract $25$ from $16$ .

$- 49 + 0 + 126 - 1 (0 - 3 \cdot - 9 + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Multiply $- 3$ by $- 9$ .

$- 49 + 0 + 126 - 1 (0 + 27 + 1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |)$

Find the determinant of $1 | \begin{array}{c} 4 & 5 \\ 2 & 1 \end{array} |$ .

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

$- 49 + 0 + 126 - 1 (0 + 27 + (4) (1) - 2 \cdot 5)$

Simplify the determinant.

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Simplify each term.

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Multiply $4$ by $1$ .

$- 49 + 0 + 126 - 1 (0 + 27 + 4 - 2 \cdot 5)$

Multiply $- 2$ by $5$ .

$- 49 + 0 + 126 - 1 (0 + 27 + 4 - 10)$

Subtract $10$ from $4$ .

$- 49 + 0 + 126 - 1 (0 + 27 - 6)$

Add $0$ and $27$ .

$- 49 + 0 + 126 - 1 (27 - 6)$

Subtract $6$ from $27$ .

$- 49 + 0 + 126 - 1 \cdot 21$

Multiply $- 1$ by $21$ .

$- 49 + 0 + 126 - 21$

Step 6

Add $- 49$ and $0$ .

$- 49 + 126 - 21$

Step 7

Add $- 49$ and $126$ .

$77 - 21$

Step 8

Subtract $21$ from $77$ .

$56$

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