Solve {ccc}{2}&{0}&{1}{-4}&{5}&{3}{1}&{2}&{9}

$$\left| \begin{array} { c c c } { 2 } & { 0 } & { 1 } \\ { - 4 } & { 5 } & { 3 } \\ { 1 } & { 2 } & { 9 } \end{array} \right|$$

Evaluate

$$65$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&0&1\\-4&5&3\\1&2&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&0&1&2&0\\-4&5&3&-4&5\\1&2&9&1&2\end{matrix}\right)$$

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

$$2\times 5\times 9-4\times 2=82$$

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

$$5+2\times 3\times 2=17$$

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

$$82-17$$

Subtract 17 from 82.

$$65$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&0&1\\-4&5&3\\1&2&9\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}5&3\\2&9\end{matrix}\right))+det(\left(\begin{matrix}-4&5\\1&2\end{matrix}\right))$$

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

$$2\left(5\times 9-2\times 3\right)-4\times 2-5$$

Simplify.

$$2\times 39-13$$

Add the terms to obtain the final result.

$$65$$

Factor

$$5\times 13$$