Find the Determinant [[-1,3,1,0],[0,0,2,0],[-2,2,-2,3],[3,-1,3,0]]

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

Step 1

Multiply every element in the $2 n d$ row by its cofactor and add.

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

Step 2

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

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

Step 3

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

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

Step 4

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

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

Step 5

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

Multiply every element in the $3 r d$ column by its cofactor and add.

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

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

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

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

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

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $- 1$ .

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

Multiply $- 3$ by $3$ .

$0 + 0 - 2 (0 - 3 (1 - 9) + 0) + 0$

Subtract $9$ from $1$ .

$0 + 0 - 2 (0 - 3 \cdot - 8 + 0) + 0$

Simplify the determinant.

Multiply $- 3$ by $- 8$ .

$0 + 0 - 2 (0 + 24 + 0) + 0$

Add $0$ and $24$ .

$0 + 0 - 2 (24 + 0) + 0$

Add $24$ and $0$ .

$0 + 0 - 2 \cdot 24 + 0$

Step 6

Simplify the determinant.

Multiply $- 2$ by $24$ .

$0 + 0 - 48 + 0$

Add $0$ and $0$ .

$0 - 48 + 0$

Subtract $48$ from $0$ .

$- 48 + 0$

Add $- 48$ and $0$ .

$- 48$

Find the Determinant [[-1,3,1,0],[0,0,2,0],[-2,2,-2,3],[3,-1,3,0]]

Related Questions

Support LCMGCF.com by disabling your adblocker.