Linear Algebra Examples

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

The determinant of $[\begin{array}{c} 0 & 1 & 2 & 1 \\ 1 & 1 & 0 & 2 \\ 1 & 0 & 1 & 0 \\ 0 & 2 & 1 & 1 \end{array}]$ is $6$ .

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Set up the determinant by breaking it into smaller components.

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

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

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

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

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Set up the determinant by breaking it into smaller components.

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

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

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

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

Simplify the determinant.

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Multiply $1$ by $1$ to get $1$ .

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

Subtract $0$ from $1$ to get $1$ .

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

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

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

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

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $0$ to get $0$ .

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

Multiply $- 1$ by $1$ to get $- 1$ .

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

Simplify the expression.

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Subtract $1$ from $0$ to get $- 1$ .

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

Multiply $2$ by $- 1$ to get $- 2$ .

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

Add $1$ and $0$ to get $1$ .

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

Subtract $2$ from $1$ to get $- 1$ .

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

Multiply $- 1$ by $- 1$ to get $1$ .

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

The determinant of $1 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{array} |$ is $5$ .

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Set up the determinant by breaking it into smaller components.

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

The determinant of $- 2 | \begin{array}{c} 1 & 2 \\ 2 & 1 \end{array} |$ is $6$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ to get $1$ .

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

Multiply $- 2$ by $2$ to get $- 4$ .

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

Simplify the expression.

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Subtract $4$ from $1$ to get $- 3$ .

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

Multiply $- 2$ by $- 3$ to get $6$ .

$0 + 1 + 6 + 0 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 1 & 2 \end{array} | + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

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

$0 + 1 + 6 + 0 - 1 | \begin{array}{c} 1 & 1 \\ 1 & 2 \end{array} | + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

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

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

$0 + 1 + 6 + 0 - 1 ((1) (2) - 1 \cdot 1) + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $1$ to get $2$ .

$0 + 1 + 6 + 0 - 1 (2 - 1 \cdot 1) + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

Multiply $- 1$ by $1$ to get $- 1$ .

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

Simplify the expression.

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Subtract $1$ from $2$ to get $1$ .

$0 + 1 + 6 + 0 - 1 \cdot 1 + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

Multiply $- 1$ by $1$ to get $- 1$ .

$0 + 1 + 6 + 0 - 1 + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

Add $6$ and $0$ to get $6$ .

$0 + 1 + 6 - 1 + 0 | \begin{array}{c} 1 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 0 \end{array} |$

Subtract $1$ from $6$ to get $5$ .

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

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

$0 + 1 + 5 + 0$

Add $0$ and $1$ to get $1$ .

$1 + 5 + 0$

Add $1$ and $5$ to get $6$ .

$6 + 0$

Add $6$ and $0$ to get $6$ .

$6$

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Linear Algebra Examples

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