Solve {ccc}{1}&{-4}&{-1}{0}&{9}&{-1}{2}&{13}&{0}

$$\left| \begin{array} { c c c } { 1 } & { - 4 } & { - 1 } \\ { 0 } & { 9 } & { - 1 } \\ { 2 } & { 13 } & { 0 } \end{array} \right|$$

Evaluate

$$39$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-4&-1\\0&9&-1\\2&13&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-4&-1&1&-4\\0&9&-1&0&9\\2&13&0&2&13\end{matrix}\right)$$

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

$$-4\left(-1\right)\times 2=8$$

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

$$2\times 9\left(-1\right)+13\left(-1\right)=-31$$

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

$$8-\left(-31\right)$$

Subtract -31 from 8.

$$39$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-4&-1\\0&9&-1\\2&13&0\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}9&-1\\13&0\end{matrix}\right))-\left(-4det(\left(\begin{matrix}0&-1\\2&0\end{matrix}\right))\right)-det(\left(\begin{matrix}0&9\\2&13\end{matrix}\right))$$

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

$$-13\left(-1\right)-\left(-4\left(-2\left(-1\right)\right)\right)-\left(-2\times 9\right)$$

Simplify.

$$13-\left(-4\times 2\right)-\left(-18\right)$$

Add the terms to obtain the final result.

$$39$$

Factor

$$3\times 13$$