Solve {ccc}{2}&{-3}&{1}{3}&{2}&{1}{-5}&{1}&{-3}=

$$\left| \begin{array} { c c c } { 2 } & { - 3 } & { 1 } \\ { 3 } & { 2 } & { 1 } \\ { - 5 } & { 1 } & { - 3 } \end{array} \right| =$$

Evaluate

$$-13$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-3&1\\3&2&1\\-5&1&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-3&1&2&-3\\3&2&1&3&2\\-5&1&-3&-5&1\end{matrix}\right)$$

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

$$2\times 2\left(-3\right)-3\left(-5\right)+3=6$$

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

$$-5\times 2+2-3\times 3\left(-3\right)=19$$

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

$$6-19$$

Subtract 19 from 6.

$$-13$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-3&1\\3&2&1\\-5&1&-3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&1\\1&-3\end{matrix}\right))-\left(-3det(\left(\begin{matrix}3&1\\-5&-3\end{matrix}\right))\right)+det(\left(\begin{matrix}3&2\\-5&1\end{matrix}\right))$$

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

$$2\left(2\left(-3\right)-1\right)-\left(-3\left(3\left(-3\right)-\left(-5\right)\right)\right)+3-\left(-5\times 2\right)$$

Simplify.

$$2\left(-7\right)-\left(-3\left(-4\right)\right)+13$$

Add the terms to obtain the final result.

$$-13$$

Factor

$$-13$$