Solve {rrr}{3}&{1}&{-5}{6}&{-7}&{-8}{23}&{24}&{25}

$$\left| \begin{array} { r r r } { 3 } & { 1 } & { - 5 } \\ { 6 } & { - 7 } & { - 8 } \\ { 23 } & { 24 } & { 25 } \end{array} \right|$$

Evaluate

$$-1808$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&1&-5\\6&-7&-8\\23&24&25\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&1&-5&3&1\\6&-7&-8&6&-7\\23&24&25&23&24\end{matrix}\right)$$

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

$$3\left(-7\right)\times 25-8\times 23-5\times 6\times 24=-1429$$

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

$$23\left(-7\right)\left(-5\right)+24\left(-8\right)\times 3+25\times 6=379$$

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

$$-1429-379$$

Subtract 379 from -1429.

$$-1808$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&1&-5\\6&-7&-8\\23&24&25\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}-7&-8\\24&25\end{matrix}\right))-det(\left(\begin{matrix}6&-8\\23&25\end{matrix}\right))-5det(\left(\begin{matrix}6&-7\\23&24\end{matrix}\right))$$

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

$$3\left(-7\times 25-24\left(-8\right)\right)-\left(6\times 25-23\left(-8\right)\right)-5\left(6\times 24-23\left(-7\right)\right)$$

Simplify.

$$3\times 17-334-5\times 305$$

Add the terms to obtain the final result.

$$-1808$$

Factor

$$-1808$$