Solve {ccc}{-3}&{1}&{-2}{22}&{-2}&{7}{4}&{-5}&{4}

$$\left| \begin{array} { c c c } { - 3 } & { 1 } & { - 2 } \\ { 22 } & { - 2 } & { 7 } \\ { 4 } & { - 5 } & { 4 } \end{array} \right|$$

Evaluate

$$63$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-3&1&-2\\22&-2&7\\4&-5&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}-3&1&-2&-3&1\\22&-2&7&22&-2\\4&-5&4&4&-5\end{matrix}\right)$$

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

$$-3\left(-2\right)\times 4+7\times 4-2\times 22\left(-5\right)=272$$

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

$$4\left(-2\right)\left(-2\right)-5\times 7\left(-3\right)+4\times 22=209$$

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

$$272-209$$

Subtract 209 from 272.

$$63$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-3&1&-2\\22&-2&7\\4&-5&4\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}-2&7\\-5&4\end{matrix}\right))-det(\left(\begin{matrix}22&7\\4&4\end{matrix}\right))-2det(\left(\begin{matrix}22&-2\\4&-5\end{matrix}\right))$$

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

$$-3\left(-2\times 4-\left(-5\times 7\right)\right)-\left(22\times 4-4\times 7\right)-2\left(22\left(-5\right)-4\left(-2\right)\right)$$

Simplify.

$$-3\times 27-60-2\left(-102\right)$$

Add the terms to obtain the final result.

$$63$$

Factor

$$3^{2}\times 7$$