Solve {ccc}{-4}&{0}&{-1}{9}&{4}&{-1}{13}&{5}&{0}

$$\left| \begin{array} { c c c } { - 4 } & { 0 } & { - 1 } \\ { 9 } & { 4 } & { - 1 } \\ { 13 } & { 5 } & { 0 } \end{array} \right|$$

Evaluate

$$-13$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-4&0&-1\\9&4&-1\\13&5&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}-4&0&-1&-4&0\\9&4&-1&9&4\\13&5&0&13&5\end{matrix}\right)$$

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

$$-9\times 5=-45$$

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

$$13\times 4\left(-1\right)+5\left(-1\right)\left(-4\right)=-32$$

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

$$-45-\left(-32\right)$$

Subtract -32 from -45.

$$-13$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-4&0&-1\\9&4&-1\\13&5&0\end{matrix}\right))$$

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.

$$-4det(\left(\begin{matrix}4&-1\\5&0\end{matrix}\right))-det(\left(\begin{matrix}9&4\\13&5\end{matrix}\right))$$

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

$$-4\left(-5\left(-1\right)\right)-\left(9\times 5-13\times 4\right)$$

Simplify.

$$-4\times 5-\left(-7\right)$$

Add the terms to obtain the final result.

$$-13$$

Factor

$$-13$$