Solve {rrr}{-1}&{2}&{3}{1}&{-3}&{0}{7}&{-2}&{2}

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

Evaluate

$$59$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-\left(-3\right)\times 2+3\left(-2\right)=0$$

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

$$7\left(-3\right)\times 3+2\times 2=-59$$

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

$$-\left(-59\right)$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-6\right)-2\times 2+3\times 19$$

Add the terms to obtain the final result.

$$59$$

Factor

$$59$$