Solve {ccc}{1}&{0}&{2}{-5}&{7}&{0}{8}&{-4}&{7}

$$\left| \begin{array} { c c c } { 1 } & { 0 } & { 2 } \\ { - 5 } & { 7 } & { 0 } \\ { 8 } & { - 4 } & { 7 } \end{array} \right|$$

Evaluate

$$-23$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&2\\-5&7&0\\8&-4&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&2&1&0\\-5&7&0&-5&7\\8&-4&7&8&-4\end{matrix}\right)$$

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

$$7\times 7+2\left(-5\right)\left(-4\right)=89$$

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

$$8\times 7\times 2=112$$

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

$$89-112$$

Subtract 112 from 89.

$$-23$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&2\\-5&7&0\\8&-4&7\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}7&0\\-4&7\end{matrix}\right))+2det(\left(\begin{matrix}-5&7\\8&-4\end{matrix}\right))$$

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

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

Simplify.

$$49+2\left(-36\right)$$

Add the terms to obtain the final result.

$$-23$$

Factor

$$-23$$