Solve {ccc}{5}&{7}&{-8}{2}&{2}&{2}{1}&{-1}&{0}

$$\left| \begin{array} { c c c } { 5 } & { 7 } & { - 8 } \\ { 2 } & { 2 } & { 2 } \\ { 1 } & { - 1 } & { 0 } \end{array} \right|$$

Evaluate

$$56$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&7&-8\\2&2&2\\1&-1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&7&-8&5&7\\2&2&2&2&2\\1&-1&0&1&-1\end{matrix}\right)$$

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

$$7\times 2-8\times 2\left(-1\right)=30$$

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

$$2\left(-8\right)-2\times 5=-26$$

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

$$30-\left(-26\right)$$

Subtract -26 from 30.

$$56$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&7&-8\\2&2&2\\1&-1&0\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}2&2\\-1&0\end{matrix}\right))-7det(\left(\begin{matrix}2&2\\1&0\end{matrix}\right))-8det(\left(\begin{matrix}2&2\\1&-1\end{matrix}\right))$$

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

$$5\left(-\left(-2\right)\right)-7\left(-2\right)-8\left(2\left(-1\right)-2\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$56$$

Factor

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