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

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

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\left(-2\right)\times 3-1=-13$$

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

$$-2+3\left(-3\right)=-11$$

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

$$-13-\left(-11\right)$$

Subtract -11 from -13.

$$-2$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-6\right)-\left(-3\times 3\right)+1$$

Add the terms to obtain the final result.

$$-2$$

Factor

$$-2$$