Solve {lll}{1}&{3}&{1}{2}&{5}&{2}{1}&{2}&{3}

$$\left| \begin{array} { l l l } { 1 } & { 3 } & { 1 } \\ { 2 } & { 5 } & { 2 } \\ { 1 } & { 2 } & { 3 } \end{array} \right|$$

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 3+3\times 2+2\times 2=25$$

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

$$5+2\times 2+3\times 2\times 3=27$$

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

$$25-27$$

Subtract 27 from 25.

$$-2$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$11-3\times 4-1$$

Add the terms to obtain the final result.

$$-2$$

Factor

$$-2$$