Find the Determinant [[4,2,-3],[5,-6,1],[0,2,-1]]

$[\begin{array}{c} 4 & 2 & - 3 \\ 5 & - 6 & 1 \\ 0 & 2 & - 1 \end{array}]$

Step 1

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

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

Step 2

Multiply $0$ by $| \begin{array}{c} 2 & - 3 \\ - 6 & 1 \end{array} |$ .

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

Step 3

Evaluate $| \begin{array}{c} - 6 & 1 \\ 2 & - 1 \end{array} |$ .

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

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

Simplify the determinant.

Simplify each term.

Multiply $- 6$ by $- 1$ .

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

Multiply $- 2$ by $1$ .

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

Subtract $2$ from $6$ .

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

Step 4

Evaluate $| \begin{array}{c} 2 & - 3 \\ 2 & - 1 \end{array} |$ .

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

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

Simplify the determinant.

Simplify each term.

Multiply $2$ by $- 1$ .

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

Multiply $- 2$ by $- 3$ .

$4 \cdot 4 - 5 (- 2 + 6) + 0$

Add $- 2$ and $6$ .

$4 \cdot 4 - 5 \cdot 4 + 0$

Step 5

Simplify the determinant.

Simplify each term.

Multiply $4$ by $4$ .

$16 - 5 \cdot 4 + 0$

Multiply $- 5$ by $4$ .

$16 - 20 + 0$

Subtract $20$ from $16$ .

$- 4 + 0$

Add $- 4$ and $0$ .

$- 4$

Find the Determinant [[4,2,-3],[5,-6,1],[0,2,-1]]

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