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

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

Evaluate

$$15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\times 3\left(-4\right)=24$$

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

$$3\times 3=9$$

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

$$24-9$$

Subtract 9 from 24.

$$15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-9-2\left(-12\right)$$

Add the terms to obtain the final result.

$$15$$

Factor

$$3\times 5$$