Find the Determinant [[1,0,-1],[0,4,-1],[6,3,0]]

$[\begin{array}{c} 1 & 0 & - 1 \\ 0 & 4 & - 1 \\ 6 & 3 & 0 \end{array}]$

Step 1

Multiply every element in the $1 s t$ row by its cofactor and add.

$1 | \begin{array}{c} 4 & - 1 \\ 3 & 0 \end{array} | + 0 | \begin{array}{c} 0 & - 1 \\ 6 & 0 \end{array} | - | \begin{array}{c} 0 & 4 \\ 6 & 3 \end{array} |$

Step 2

Multiply $0$ by $| \begin{array}{c} 0 & - 1 \\ 6 & 0 \end{array} |$ .

$1 | \begin{array}{c} 4 & - 1 \\ 3 & 0 \end{array} | + 0 - | \begin{array}{c} 0 & 4 \\ 6 & 3 \end{array} |$

Step 3

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $4$ by $0$ .

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

Multiply $- 3$ by $- 1$ .

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

Add $0$ and $3$ .

$1 \cdot 3 + 0 - | \begin{array}{c} 0 & 4 \\ 6 & 3 \end{array} |$

Step 4

Evaluate $| \begin{array}{c} 0 & 4 \\ 6 & 3 \end{array} |$ .

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

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

Simplify the determinant.

Simplify each term.

Multiply $0$ by $3$ .

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

Multiply $- 6$ by $4$ .

$1 \cdot 3 + 0 - (0 - 24)$

Subtract $24$ from $0$ .

$1 \cdot 3 + 0 - - 24$

Step 5

Simplify the determinant.

Simplify each term.

Multiply $3$ by $1$ .

$3 + 0 - - 24$

Multiply $- 1$ by $- 24$ .

$3 + 0 + 24$

Add $3$ and $0$ .

$3 + 24$

Add $3$ and $24$ .

$27$

Find the Determinant [[1,0,-1],[0,4,-1],[6,3,0]]

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