Solve {ccc}{4}&{2}&{2}{1}&{-5}&{3}{8}&{7}&{-1}

$$\left| \begin{array} { c c c } { 4 } & { 2 } & { 2 } \\ { 1 } & { - 5 } & { 3 } \\ { 8 } & { 7 } & { - 1 } \end{array} \right|$$

Evaluate

$$80$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&2&2\\1&-5&3\\8&7&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&2&2&4&2\\1&-5&3&1&-5\\8&7&-1&8&7\end{matrix}\right)$$

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

$$4\left(-5\right)\left(-1\right)+2\times 3\times 8+2\times 7=82$$

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

$$8\left(-5\right)\times 2+7\times 3\times 4-2=2$$

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

$$82-2$$

Subtract 2 from 82.

$$80$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&2&2\\1&-5&3\\8&7&-1\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}-5&3\\7&-1\end{matrix}\right))-2det(\left(\begin{matrix}1&3\\8&-1\end{matrix}\right))+2det(\left(\begin{matrix}1&-5\\8&7\end{matrix}\right))$$

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

$$4\left(-5\left(-1\right)-7\times 3\right)-2\left(-1-8\times 3\right)+2\left(7-8\left(-5\right)\right)$$

Simplify.

$$4\left(-16\right)-2\left(-25\right)+2\times 47$$

Add the terms to obtain the final result.

$$80$$

Factor

$$2^{4}\times 5$$