Solve {ccc}{2}&{1}&{-5}{1}&{-3}&{0}{1}&{4}&{-7}

$$\left| \begin{array} { c c c } { 2 } & { 1 } & { - 5 } \\ { 1 } & { - 3 } & { 0 } \\ { 1 } & { 4 } & { - 7 } \end{array} \right|$$

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&1&-5\\1&-3&0\\1&4&-7\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&1&-5&2&1\\1&-3&0&1&-3\\1&4&-7&1&4\end{matrix}\right)$$

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

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

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

$$-3\left(-5\right)-7=8$$

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

$$22-8$$

Subtract 8 from 22.

$$14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&1&-5\\1&-3&0\\1&4&-7\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}-3&0\\4&-7\end{matrix}\right))-det(\left(\begin{matrix}1&0\\1&-7\end{matrix}\right))-5det(\left(\begin{matrix}1&-3\\1&4\end{matrix}\right))$$

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

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

Simplify.

$$2\times 21-\left(-7\right)-5\times 7$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$