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

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

Set up the determinant by breaking it into smaller components.

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

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $3$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify the expression.

Subtract $1$ from $3$ .

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

Multiply $- 2$ by $2$ .

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

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $1$ by $1$ .

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

Multiply $- 1$ by $- 2$ .

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

Simplify the expression.

Add $1$ and $2$ .

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

Multiply $- 1$ by $3$ .

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

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

$- 4 - 3 + 0$

Subtract $3$ from $- 4$ .

$- 7 + 0$

Add $- 7$ and $0$ .

$- 7$

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

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