Algebra Examples

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

Step 1

Refer to the corresponding sign matrix below.

$[\begin{array}{c} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{array}]$

Step 2

Use the sign matrix and the given matrix, $A$ , to find the cofactor of each element.

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Find the determinant of $[\begin{array}{c} 0 & x & 4 \\ 0 & 1 & - 1 \\ - 1 & 2 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{11} = 0 | \begin{array}{c} 1 & - 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |$

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

$A_{11} = 0 + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |$

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

$A_{11} = 0 + 0 - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |$

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

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

$A_{11} = 0 + 0 - 1 ((x) (- 1) - 1 \cdot 4)$

Simplify the determinant.

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Simplify each term.

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Move $- 1$ to the left of $x$ .

$A_{11} = 0 + 0 - 1 (- 1 x - 1 \cdot 4)$

Rewrite $- 1 x$ as $- x$ .

$A_{11} = 0 + 0 - 1 (- x - 1 \cdot 4)$

Multiply $- 1$ by $4$ .

$A_{11} = 0 + 0 - 1 (- x - 4)$

Apply the distributive property.

$A_{11} = 0 + 0 - 1 (- x) - 1 \cdot - 4$

Multiply $- 1 (- x)$ .

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Multiply $- 1$ by $- 1$ .

$A_{11} = 0 + 0 + 1 x - 1 \cdot - 4$

Multiply $x$ by $1$ .

$A_{11} = 0 + 0 + x - 1 \cdot - 4$

Multiply $- 1$ by $- 4$ .

$A_{11} = 0 + 0 + x + 4$

Combine the opposite terms in $0 + 0 + x + 4$ .

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Add $0$ and $0$ .

$A_{11} = 0 + x + 4$

Add $0$ and $x$ .

$A_{11} = x + 4$

Find the determinant of $- [\begin{array}{c} 4 & x & 4 \\ 0 & 1 & - 1 \\ - 1 & 2 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{12} = - (4 | \begin{array}{c} 1 & - 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

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

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

$A_{12} = - (4 ((1) (0) - 2 \cdot - 1) + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $0$ by $1$ .

$A_{12} = - (4 (0 - 2 \cdot - 1) + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

Multiply $- 2$ by $- 1$ .

$A_{12} = - (4 (0 + 2) + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

Simplify the expression.

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Add $0$ and $2$ .

$A_{12} = - (4 \cdot 2 + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

Multiply $4$ by $2$ .

$A_{12} = - (8 + 0 | \begin{array}{c} x & 4 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

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

$A_{12} = - (8 + 0 - 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} |)$

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

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

$A_{12} = - (8 + 0 - 1 ((x) (- 1) - 1 \cdot 4))$

Simplify the determinant.

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Simplify each term.

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Move $- 1$ to the left of $x$ .

$A_{12} = - (8 + 0 - 1 (- 1 x - 1 \cdot 4))$

Rewrite $- 1 x$ as $- x$ .

$A_{12} = - (8 + 0 - 1 (- x - 1 \cdot 4))$

Multiply $- 1$ by $4$ .

$A_{12} = - (8 + 0 - 1 (- x - 4))$

Apply the distributive property.

$A_{12} = - (8 + 0 - 1 (- x) - 1 \cdot - 4)$

Multiply $- 1 (- x)$ .

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Multiply $- 1$ by $- 1$ .

$A_{12} = - (8 + 0 + 1 x - 1 \cdot - 4)$

Multiply $x$ by $1$ .

$A_{12} = - (8 + 0 + x - 1 \cdot - 4)$

Multiply $- 1$ by $- 4$ .

$A_{12} = - (8 + 0 + x + 4)$

Add $8$ and $0$ .

$A_{12} = - (8 + x + 4)$

Add $8$ and $4$ .

$A_{12} = - (x + 12)$

Simplify the determinant.

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Apply the distributive property.

$A_{12} = - x - 1 \cdot 12$

Multiply $- 1$ by $12$ .

$A_{12} = - x - 12$

Find the determinant of $[\begin{array}{c} 4 & 0 & 4 \\ 0 & 0 & - 1 \\ - 1 & - 1 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{13} = 0 | \begin{array}{c} 0 & - 1 \\ - 1 & 0 \end{array} | + 0 | \begin{array}{c} 4 & 4 \\ - 1 & 0 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 0 & - 1 \end{array} |$

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

$A_{13} = 0 + 0 | \begin{array}{c} 4 & 4 \\ - 1 & 0 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 0 & - 1 \end{array} |$

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

$A_{13} = 0 + 0 + 1 | \begin{array}{c} 4 & 4 \\ 0 & - 1 \end{array} |$

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

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

$A_{13} = 0 + 0 + 1 ((4) (- 1) + 0 \cdot 4)$

Simplify the determinant.

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Multiply $(4) (- 1) + 0 \cdot 4$ by $1$ .

$A_{13} = 0 + 0 + (4) (- 1) + 0 \cdot 4$

Simplify each term.

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Multiply $4$ by $- 1$ .

$A_{13} = 0 + 0 - 4 + 0 \cdot 4$

Multiply $0$ by $4$ .

$A_{13} = 0 + 0 - 4 + 0$

Add $- 4$ and $0$ .

$A_{13} = 0 + 0 - 4$

Add $0$ and $0$ .

$A_{13} = 0 - 4$

Subtract $4$ from $0$ .

$A_{13} = - 4$

Find the determinant of $- [\begin{array}{c} 4 & 0 & x \\ 0 & 0 & 1 \\ - 1 & - 1 & 2 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{14} = - (0 | \begin{array}{c} 0 & 1 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} 4 & x \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 4 & x \\ 0 & 1 \end{array} |)$

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

$A_{14} = - (0 + 0 | \begin{array}{c} 4 & x \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 4 & x \\ 0 & 1 \end{array} |)$

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

$A_{14} = - (0 + 0 + 1 | \begin{array}{c} 4 & x \\ 0 & 1 \end{array} |)$

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

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

$A_{14} = - (0 + 0 + 1 ((4) (1) + 0 x))$

Simplify the determinant.

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Multiply $(4) (1) + 0 x$ by $1$ .

$A_{14} = - (0 + 0 + (4) (1) + 0 x)$

Simplify each term.

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Multiply $4$ by $1$ .

$A_{14} = - (0 + 0 + 4 + 0 x)$

Multiply $0$ by $x$ .

$A_{14} = - (0 + 0 + 4 + 0)$

Add $4$ and $0$ .

$A_{14} = - (0 + 0 + 4)$

Add $0$ and $0$ .

$A_{14} = - (0 + 4)$

Add $0$ and $4$ .

$A_{14} = - (4)$

Multiply $- 1$ by $4$ .

$A_{14} = - 4$

Find the determinant of $- [\begin{array}{c} - 1 & 3 & 0 \\ 0 & 1 & - 1 \\ - 1 & 2 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{21} = - (0 | \begin{array}{c} 0 & 1 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

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

$A_{21} = - (0 + 1 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

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

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

$A_{21} = - (0 + 1 ((- 1) (2) + 1 \cdot 3) + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

Simplify the determinant.

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Multiply $(- 1) (2) + 1 \cdot 3$ by $1$ .

$A_{21} = - (0 + (- 1) (2) + 1 \cdot 3 + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

Simplify each term.

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Multiply $- 1$ by $2$ .

$A_{21} = - (0 - 2 + 1 \cdot 3 + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

Multiply $3$ by $1$ .

$A_{21} = - (0 - 2 + 3 + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

Add $- 2$ and $3$ .

$A_{21} = - (0 + 1 + 0 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |)$

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

$A_{21} = - (0 + 1 + 0)$

Add $0$ and $1$ .

$A_{21} = - (1 + 0)$

Add $1$ and $0$ .

$A_{21} = - (1)$

Multiply $- 1$ by $1$ .

$A_{21} = - 1$

Find the determinant of $[\begin{array}{c} 2 & 3 & 0 \\ 0 & 1 & - 1 \\ - 1 & 2 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{22} = 0 | \begin{array}{c} 0 & 1 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

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

$A_{22} = 0 + 1 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

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

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

$A_{22} = 0 + 1 ((2) (2) + 1 \cdot 3) + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

Simplify the determinant.

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Multiply $(2) (2) + 1 \cdot 3$ by $1$ .

$A_{22} = 0 + (2) (2) + 1 \cdot 3 + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

Simplify each term.

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Multiply $2$ by $2$ .

$A_{22} = 0 + 4 + 1 \cdot 3 + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

Multiply $3$ by $1$ .

$A_{22} = 0 + 4 + 3 + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

Add $4$ and $3$ .

$A_{22} = 0 + 7 + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} |$

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

$A_{22} = 0 + 7 + 0$

Add $0$ and $7$ .

$A_{22} = 7 + 0$

Add $7$ and $0$ .

$A_{22} = 7$

Find the determinant of $- [\begin{array}{c} 2 & - 1 & 0 \\ 0 & 0 & - 1 \\ - 1 & - 1 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{23} = - (0 | \begin{array}{c} 0 & 0 \\ - 1 & - 1 \end{array} | + 1 | \begin{array}{c} 2 & - 1 \\ - 1 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

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

$A_{23} = - (0 + 1 | \begin{array}{c} 2 & - 1 \\ - 1 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

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

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

$A_{23} = - (0 + 1 ((2) (- 1) + 1 \cdot - 1) + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

Simplify the determinant.

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Multiply $(2) (- 1) + 1 \cdot - 1$ by $1$ .

$A_{23} = - (0 + (2) (- 1) + 1 \cdot - 1 + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

Simplify each term.

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Multiply $2$ by $- 1$ .

$A_{23} = - (0 - 2 + 1 \cdot - 1 + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

Multiply $- 1$ by $1$ .

$A_{23} = - (0 - 2 - 1 + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

Subtract $1$ from $- 2$ .

$A_{23} = - (0 - 3 + 0 | \begin{array}{c} 2 & - 1 \\ 0 & 0 \end{array} |)$

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

$A_{23} = - (0 - 3 + 0)$

Subtract $3$ from $0$ .

$A_{23} = - (- 3 + 0)$

Add $- 3$ and $0$ .

$A_{23} = - (- 3)$

Multiply $- 1$ by $- 3$ .

$A_{23} = 3$

Find the determinant of $[\begin{array}{c} 2 & - 1 & 3 \\ 0 & 0 & 1 \\ - 1 & - 1 & 2 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{24} = 2 | \begin{array}{c} 0 & 1 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

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

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

$A_{24} = 2 ((0) (2) + 1 \cdot 1) + 0 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $0$ by $2$ .

$A_{24} = 2 (0 + 1 \cdot 1) + 0 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

Multiply $1$ by $1$ .

$A_{24} = 2 (0 + 1) + 0 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

Simplify the expression.

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Add $0$ and $1$ .

$A_{24} = 2 \cdot 1 + 0 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

Multiply $2$ by $1$ .

$A_{24} = 2 + 0 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

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

$A_{24} = 2 + 0 - 1 | \begin{array}{c} - 1 & 3 \\ 0 & 1 \end{array} |$

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

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

$A_{24} = 2 + 0 - 1 ((- 1) (1) + 0 \cdot 3)$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $1$ .

$A_{24} = 2 + 0 - 1 (- 1 + 0 \cdot 3)$

Multiply $0$ by $3$ .

$A_{24} = 2 + 0 - 1 (- 1 + 0)$

Simplify the expression.

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Add $- 1$ and $0$ .

$A_{24} = 2 + 0 - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$A_{24} = 2 + 0 + 1$

Add $2$ and $0$ .

$A_{24} = 2 + 1$

Add $2$ and $1$ .

$A_{24} = 3$

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

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Set up the determinant by breaking it into smaller components.

$A_{31} = 0 | \begin{array}{c} 0 & x \\ - 1 & 2 \end{array} | - 4 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

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

$A_{31} = 0 - 4 | \begin{array}{c} - 1 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

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

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

$A_{31} = 0 - 4 ((- 1) (2) + 1 \cdot 3) + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $2$ .

$A_{31} = 0 - 4 (- 2 + 1 \cdot 3) + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

Multiply $3$ by $1$ .

$A_{31} = 0 - 4 (- 2 + 3) + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

Simplify the expression.

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Add $- 2$ and $3$ .

$A_{31} = 0 - 4 \cdot 1 + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

Multiply $- 4$ by $1$ .

$A_{31} = 0 - 4 + 0 | \begin{array}{c} - 1 & 3 \\ 0 & x \end{array} |$

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

$A_{31} = 0 - 4 + 0$

Subtract $4$ from $0$ .

$A_{31} = - 4 + 0$

Add $- 4$ and $0$ .

$A_{31} = - 4$

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

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Set up the determinant by breaking it into smaller components.

$A_{32} = - (0 | \begin{array}{c} 4 & x \\ - 1 & 2 \end{array} | - 4 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

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

$A_{32} = - (0 - 4 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

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

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

$A_{32} = - (0 - 4 ((2) (2) + 1 \cdot 3) + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $2$ .

$A_{32} = - (0 - 4 (4 + 1 \cdot 3) + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Multiply $3$ by $1$ .

$A_{32} = - (0 - 4 (4 + 3) + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Simplify the expression.

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Add $4$ and $3$ .

$A_{32} = - (0 - 4 \cdot 7 + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Multiply $- 4$ by $7$ .

$A_{32} = - (0 - 28 + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

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

$A_{32} = - (0 - 28 + 0)$

Subtract $28$ from $0$ .

$A_{32} = - (- 28 + 0)$

Add $- 28$ and $0$ .

$A_{32} = - (- 28)$

Multiply $- 1$ by $- 28$ .

$A_{32} = 28$

Find the determinant of $[\begin{array}{c} 2 & - 1 & 0 \\ 4 & 0 & 4 \\ - 1 & - 1 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{33} = 0 | \begin{array}{c} 4 & 0 \\ - 1 & - 1 \end{array} | - 4 | \begin{array}{c} 2 & - 1 \\ - 1 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

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

$A_{33} = 0 - 4 | \begin{array}{c} 2 & - 1 \\ - 1 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

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

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

$A_{33} = 0 - 4 ((2) (- 1) + 1 \cdot - 1) + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $- 1$ .

$A_{33} = 0 - 4 (- 2 + 1 \cdot - 1) + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

Multiply $- 1$ by $1$ .

$A_{33} = 0 - 4 (- 2 - 1) + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

Simplify the expression.

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Subtract $1$ from $- 2$ .

$A_{33} = 0 - 4 \cdot - 3 + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

Multiply $- 4$ by $- 3$ .

$A_{33} = 0 + 12 + 0 | \begin{array}{c} 2 & - 1 \\ 4 & 0 \end{array} |$

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

$A_{33} = 0 + 12 + 0$

Add $0$ and $12$ .

$A_{33} = 12 + 0$

Add $12$ and $0$ .

$A_{33} = 12$

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

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Set up the determinant by breaking it into smaller components.

$A_{34} = - (1 | \begin{array}{c} 4 & x \\ - 1 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

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

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

$A_{34} = - (1 ((4) (2) + 1 x) + 0 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Simplify the determinant.

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Multiply $(4) (2) + 1 x$ by $1$ .

$A_{34} = - ((4) (2) + 1 x + 0 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Simplify each term.

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Multiply $4$ by $2$ .

$A_{34} = - (8 + 1 x + 0 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

Multiply $x$ by $1$ .

$A_{34} = - (8 + x + 0 | \begin{array}{c} 2 & 3 \\ - 1 & 2 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

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

$A_{34} = - (8 + x + 0 + 1 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |)$

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

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

$A_{34} = - (8 + x + 0 + 1 ((2) (x) - 4 \cdot 3))$

Simplify the determinant.

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Multiply $(2) (x) - 4 \cdot 3$ by $1$ .

$A_{34} = - (8 + x + 0 + (2) (x) - 4 \cdot 3)$

Multiply $- 4$ by $3$ .

$A_{34} = - (8 + x + 0 + 2 x - 12)$

Simplify by subtracting numbers.

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Add $8 + x$ and $0$ .

$A_{34} = - (8 + x + 2 x - 12)$

Subtract $12$ from $8$ .

$A_{34} = - (x + 2 x - 4)$

Add $x$ and $2 x$ .

$A_{34} = - (3 x - 4)$

Simplify the determinant.

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Apply the distributive property.

$A_{34} = - (3 x) + 4$

Multiply.

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Multiply $3$ by $- 1$ .

$A_{34} = - 3 x + 4$

Multiply $- 1$ by $- 4$ .

$A_{34} = - 3 x + 4$

Find the determinant of $- [\begin{array}{c} - 1 & 3 & 0 \\ 0 & x & 4 \\ 0 & 1 & - 1 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{41} = - (- 1 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

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

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

$A_{41} = - (- 1 ((x) (- 1) - 1 \cdot 4) + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Simplify the determinant.

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Simplify each term.

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Move $- 1$ to the left of $x$ .

$A_{41} = - (- 1 (- 1 x - 1 \cdot 4) + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Rewrite $- 1 x$ as $- x$ .

$A_{41} = - (- 1 (- x - 1 \cdot 4) + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Multiply $- 1$ by $4$ .

$A_{41} = - (- 1 (- x - 4) + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Apply the distributive property.

$A_{41} = - (- 1 (- x) - 1 \cdot - 4 + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Multiply $- 1 (- x)$ .

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Multiply $- 1$ by $- 1$ .

$A_{41} = - (1 x - 1 \cdot - 4 + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Multiply $x$ by $1$ .

$A_{41} = - (x - 1 \cdot - 4 + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

Multiply $- 1$ by $- 4$ .

$A_{41} = - (x + 4 + 0 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

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

$A_{41} = - (x + 4 + 0 + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |)$

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

$A_{41} = - (x + 4 + 0 + 0)$

Combine the opposite terms in $x + 4 + 0 + 0$ .

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Add $x + 4$ and $0$ .

$A_{41} = - (x + 4 + 0)$

Add $x + 4$ and $0$ .

$A_{41} = - (x + 4)$

Simplify the determinant.

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Apply the distributive property.

$A_{41} = - x - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$A_{41} = - x - 4$

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

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Set up the determinant by breaking it into smaller components.

$A_{42} = 2 | \begin{array}{c} x & 4 \\ 1 & - 1 \end{array} | - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

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

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

$A_{42} = 2 ((x) (- 1) - 1 \cdot 4) - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Move $- 1$ to the left of $x$ .

$A_{42} = 2 (- 1 x - 1 \cdot 4) - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Rewrite $- 1 x$ as $- x$ .

$A_{42} = 2 (- x - 1 \cdot 4) - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Multiply $- 1$ by $4$ .

$A_{42} = 2 (- x - 4) - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Simplify by multiplying through.

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Apply the distributive property.

$A_{42} = 2 (- x) + 2 \cdot - 4 - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Multiply.

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Multiply $- 1$ by $2$ .

$A_{42} = - 2 x + 2 \cdot - 4 - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Multiply $2$ by $- 4$ .

$A_{42} = - 2 x - 8 - 4 | \begin{array}{c} 3 & 0 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

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

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

$A_{42} = - 2 x - 8 - 4 ((3) (- 1) - 1 \cdot 0) + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Simplify the determinant.

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Multiply $3$ by $- 1$ .

$A_{42} = - 2 x - 8 - 4 (- 3 - 1 \cdot 0) + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Subtract $0$ from $- 3$ .

$A_{42} = - 2 x - 8 - 4 \cdot - 3 + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

Multiply $- 4$ by $- 3$ .

$A_{42} = - 2 x - 8 + 12 + 0 | \begin{array}{c} 3 & 0 \\ x & 4 \end{array} |$

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

$A_{42} = - 2 x - 8 + 12 + 0$

Add $- 2 x - 8 + 12$ and $0$ .

$A_{42} = - 2 x - 8 + 12$

Add $- 8$ and $12$ .

$A_{42} = - 2 x + 4$

Find the determinant of $- [\begin{array}{c} 2 & - 1 & 0 \\ 4 & 0 & 4 \\ 0 & 0 & - 1 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$A_{43} = - (1 | \begin{array}{c} 4 & 4 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

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

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

$A_{43} = - (1 ((4) (- 1) + 0 \cdot 4) + 0 | \begin{array}{c} 2 & 0 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

Simplify the determinant.

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Multiply $(4) (- 1) + 0 \cdot 4$ by $1$ .

$A_{43} = - ((4) (- 1) + 0 \cdot 4 + 0 | \begin{array}{c} 2 & 0 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

Simplify each term.

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Multiply $4$ by $- 1$ .

$A_{43} = - (- 4 + 0 \cdot 4 + 0 | \begin{array}{c} 2 & 0 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

Multiply $0$ by $4$ .

$A_{43} = - (- 4 + 0 + 0 | \begin{array}{c} 2 & 0 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

Add $- 4$ and $0$ .

$A_{43} = - (- 4 + 0 | \begin{array}{c} 2 & 0 \\ 0 & - 1 \end{array} | + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

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

$A_{43} = - (- 4 + 0 + 0 | \begin{array}{c} 2 & 0 \\ 4 & 4 \end{array} |)$

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

$A_{43} = - (- 4 + 0 + 0)$

Add $- 4$ and $0$ .

$A_{43} = - (- 4 + 0)$

Add $- 4$ and $0$ .

$A_{43} = - (- 4)$

Multiply $- 1$ by $- 4$ .

$A_{43} = 4$

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

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Set up the determinant by breaking it into smaller components.

$A_{44} = 1 | \begin{array}{c} 4 & x \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

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

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

$A_{44} = 1 ((4) (1) + 0 x) + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

Simplify the determinant.

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Multiply $(4) (1) + 0 x$ by $1$ .

$A_{44} = (4) (1) + 0 x + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

Simplify each term.

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Multiply $4$ by $1$ .

$A_{44} = 4 + 0 x + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

Multiply $0$ by $x$ .

$A_{44} = 4 + 0 + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

Add $4$ and $0$ .

$A_{44} = 4 + 0 | \begin{array}{c} 2 & 3 \\ 0 & 1 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

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

$A_{44} = 4 + 0 + 0 | \begin{array}{c} 2 & 3 \\ 4 & x \end{array} |$

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

$A_{44} = 4 + 0 + 0$

Add $4$ and $0$ .

$A_{44} = 4 + 0$

Add $4$ and $0$ .

$A_{44} = 4$

The cofactor matrix is a matrix with cofactors as the elements.

$[\begin{array}{c} x + 4 & - x - 12 & - 4 & - 4 \\ - 1 & 7 & 3 & 3 \\ - 4 & 28 & 12 & - 3 x + 4 \\ - x - 4 & - 2 x + 4 & 4 & 4 \end{array}]$