Solve {lll}{1}&{2}&{-2}{0}&{-5}&{2}{3}&{-4}&{0}

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

Evaluate

$$-10$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2\times 3=12$$

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

$$3\left(-5\right)\left(-2\right)-4\times 2=22$$

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

$$12-22$$

Subtract 22 from 12.

$$-10$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$8-2\left(-6\right)-2\times 15$$

Add the terms to obtain the final result.

$$-10$$

Factor

$$-10$$