Solve {ccc}{2}&{5}&{3}{1}&{4}&{-5}{5}&{3}&{5}

$$\left| \begin{array} { c c c } { 2 } & { 5 } & { 3 } \\ { 1 } & { 4 } & { - 5 } \\ { 5 } & { 3 } & { 5 } \end{array} \right|$$

Evaluate

$$-131$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&5&3\\1&4&-5\\5&3&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&5&3&2&5\\1&4&-5&1&4\\5&3&5&5&3\end{matrix}\right)$$

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

$$2\times 4\times 5+5\left(-5\right)\times 5+3\times 3=-76$$

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

$$5\times 4\times 3+3\left(-5\right)\times 2+5\times 5=55$$

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

$$-76-55$$

Subtract 55 from -76.

$$-131$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&5&3\\1&4&-5\\5&3&5\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}4&-5\\3&5\end{matrix}\right))-5det(\left(\begin{matrix}1&-5\\5&5\end{matrix}\right))+3det(\left(\begin{matrix}1&4\\5&3\end{matrix}\right))$$

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

$$2\left(4\times 5-3\left(-5\right)\right)-5\left(5-5\left(-5\right)\right)+3\left(3-5\times 4\right)$$

Simplify.

$$2\times 35-5\times 30+3\left(-17\right)$$

Add the terms to obtain the final result.

$$-131$$

Factor

$$-131$$