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

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

Evaluate

$$96$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 5\times 2+3\left(-4\right)+4\times 2\times 2=44$$

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

$$-4\times 5\times 4+2\times 3\times 4+2\times 2=-52$$

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

$$44-\left(-52\right)$$

Subtract -52 from 44.

$$96$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$4\times 4-16+4\times 24$$

Add the terms to obtain the final result.

$$96$$

Factor

$$2^{5}\times 3$$