Solve {ccc}{-3}&{2}&{-2}{-1}&{1}&{2}{0}&{3}&{-1}

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

Calculate Determinant

$$25$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}-3&2&-2\\-1&1&2\\0&3&-1\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}-3&2&-2&-3&2\\-1&1&2&-1&1\\0&3&-1&0&3\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$-3\left(-1\right)-2\left(-1\right)\times 3=9$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$3\times 2\left(-3\right)-\left(-2\right)=-16$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$9-\left(-16\right)$$

Subtract -16 from 9.

$$25$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}-3&2&-2\\-1&1&2\\0&3&-1\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$-3det(\left(\begin{matrix}1&2\\3&-1\end{matrix}\right))-2det(\left(\begin{matrix}-1&2\\0&-1\end{matrix}\right))-2det(\left(\begin{matrix}-1&1\\0&3\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$-3\left(-1-3\times 2\right)-2\left(-1\right)\left(-1\right)-2\left(-1\right)\times 3$$

Simplify.

$$-3\left(-7\right)-2-2\left(-3\right)$$

Add the terms to obtain the final result.

$$25$$

Evaluate

$$\left(\begin{matrix}-3&2&-2\\-1&1&2\\0&3&-1\end{matrix}\right)$$