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

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

Evaluate

$$62$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\times 3\left(-2\right)+7\times 4\times 2=74$$

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

$$-4\left(-3\right)=12$$

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

$$74-12$$

Subtract 12 from 74.

$$62$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$62$$

Factor

$$2\times 31$$