Linear Algebra Examples

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

Set up the determinant by breaking it into smaller components.

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

The determinant of $1 | \begin{array}{c} 2 & 1 & 4 \\ 1 & 2 & 3 \\ 1 & 1 & 0 \end{array} |$ is $- 7$ .

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Set up the determinant by breaking it into smaller components.

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

The determinant of $4 | \begin{array}{c} 1 & 2 \\ 1 & 1 \end{array} |$ is $- 4$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Simplify the expression.

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Subtract $2$ from $1$ .

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

Multiply $4$ by $- 1$ .

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

The determinant of $- 3 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |$ is $- 3$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify the expression.

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Subtract $1$ from $2$ .

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

Multiply $- 3$ by $1$ .

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

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

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

Subtract $3$ from $- 4$ .

$- 7 + 0 + 0 | \begin{array}{c} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 1 & 0 \end{array} | + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

Add $- 7$ and $0$ .

$- 7 + 0 | \begin{array}{c} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 1 & 0 \end{array} | + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

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

$- 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

The determinant of $2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} |$ is $14$ .

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Set up the determinant by breaking it into smaller components.

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

The determinant of $3 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |$ is $3$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify the expression.

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Subtract $1$ from $2$ .

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

Multiply $3$ by $1$ .

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

The determinant of $- 4 | \begin{array}{c} 1 & 2 \\ 1 & 1 \end{array} |$ is $4$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Simplify the expression.

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Subtract $2$ from $1$ .

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

Multiply $- 4$ by $- 1$ .

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

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

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

Add $3$ and $4$ .

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

Add $7$ and $0$ .

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

Multiply $2$ by $7$ .

$- 7 + 0 + 14 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

The determinant of $- 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$ is $0$ .

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Set up the determinant by breaking it into smaller components.

$- 7 + 0 + 14 - 2 (1 | \begin{array}{c} 1 & 4 \\ 2 & 3 \end{array} | - 2 | \begin{array}{c} 2 & 3 \\ 2 & 3 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

The determinant of $1 | \begin{array}{c} 1 & 4 \\ 2 & 3 \end{array} |$ is $- 5$ .

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

$- 7 + 0 + 14 - 2 ((1) (3) - 2 \cdot 4 - 2 | \begin{array}{c} 2 & 3 \\ 2 & 3 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $1$ .

$- 7 + 0 + 14 - 2 (3 - 2 \cdot 4 - 2 | \begin{array}{c} 2 & 3 \\ 2 & 3 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Multiply $- 2$ by $4$ .

$- 7 + 0 + 14 - 2 (3 - 8 - 2 | \begin{array}{c} 2 & 3 \\ 2 & 3 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Subtract $8$ from $3$ .

$- 7 + 0 + 14 - 2 (- 5 - 2 | \begin{array}{c} 2 & 3 \\ 2 & 3 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

The determinant of $- 2 | \begin{array}{c} 2 & 3 \\ 2 & 3 \end{array} |$ is $0$ .

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

$- 7 + 0 + 14 - 2 (- 5 - 2 ((2) (3) - 2 \cdot 3) + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $3$ .

$- 7 + 0 + 14 - 2 (- 5 - 2 (6 - 2 \cdot 3) + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Multiply $- 2$ by $3$ .

$- 7 + 0 + 14 - 2 (- 5 - 2 (6 - 6) + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Simplify the expression.

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Subtract $6$ from $6$ .

$- 7 + 0 + 14 - 2 (- 5 - 2 \cdot 0 + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

Multiply $- 2$ by $0$ .

$- 7 + 0 + 14 - 2 (- 5 + 0 + 1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |)$

The determinant of $1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} |$ is $5$ .

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

$- 7 + 0 + 14 - 2 (- 5 + 0 + (2) (4) - 1 \cdot 3)$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $4$ .

$- 7 + 0 + 14 - 2 (- 5 + 0 + 8 - 1 \cdot 3)$

Multiply $- 1$ by $3$ .

$- 7 + 0 + 14 - 2 (- 5 + 0 + 8 - 3)$

Subtract $3$ from $8$ .

$- 7 + 0 + 14 - 2 (- 5 + 0 + 5)$

Add $- 5$ and $0$ .

$- 7 + 0 + 14 - 2 (- 5 + 5)$

Add $- 5$ and $5$ .

$- 7 + 0 + 14 - 2 \cdot 0$

Multiply $- 2$ by $0$ .

$- 7 + 0 + 14 + 0$

Add $- 7$ and $0$ .

$- 7 + 14 + 0$

Add $- 7$ and $14$ .

$7 + 0$

Add $7$ and $0$ .

$7$

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