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

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

Evaluate

$$-8$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\left(-3\right)=-9$$

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

$$-1=-1$$

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

$$-9-\left(-1\right)$$

Subtract -1 from -9.

$$-8$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-8$$

Factor

$$-8$$