Solve {lll}{2}&{7}&{5}{1}&{2}&{3}{2}&{-1}&{3}

$$\left| \begin{array} { l l l } { 2 } & { 7 } & { 5 } \\ { 1 } & { 2 } & { 3 } \\ { 2 } & { - 1 } & { 3 } \end{array} \right|$$

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&7&5\\1&2&3\\2&-1&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&7&5&2&7\\1&2&3&1&2\\2&-1&3&2&-1\end{matrix}\right)$$

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

$$2\times 2\times 3+7\times 3\times 2+5\left(-1\right)=49$$

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

$$2\times 2\times 5-3\times 2+3\times 7=35$$

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

$$49-35$$

Subtract 35 from 49.

$$14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&7&5\\1&2&3\\2&-1&3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&3\\-1&3\end{matrix}\right))-7det(\left(\begin{matrix}1&3\\2&3\end{matrix}\right))+5det(\left(\begin{matrix}1&2\\2&-1\end{matrix}\right))$$

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

$$2\left(2\times 3-\left(-3\right)\right)-7\left(3-2\times 3\right)+5\left(-1-2\times 2\right)$$

Simplify.

$$2\times 9-7\left(-3\right)+5\left(-5\right)$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$