Solve {lll}{1}&{0}&{-4}{0}&{4}&{5}{2}&{5}&{13}=

$$\left| \begin{array} { l l l } { 1 } & { 0 } & { - 4 } \\ { 0 } & { 4 } & { 5 } \\ { 2 } & { 5 } & { 13 } \end{array} \right| =$$

Evaluate

$$59$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&-4\\0&4&5\\2&5&13\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&-4&1&0\\0&4&5&0&4\\2&5&13&2&5\end{matrix}\right)$$

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

$$4\times 13=52$$

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

$$2\times 4\left(-4\right)+5\times 5=-7$$

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

$$52-\left(-7\right)$$

Subtract -7 from 52.

$$59$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&-4\\0&4&5\\2&5&13\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&5\\5&13\end{matrix}\right))-4det(\left(\begin{matrix}0&4\\2&5\end{matrix}\right))$$

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

$$4\times 13-5\times 5-4\left(-2\times 4\right)$$

Simplify.

$$27-4\left(-8\right)$$

Add the terms to obtain the final result.

$$59$$

Factor

$$59$$