Solve {rrr}{5}&{6}&{3}{0}&{-2}&{1}{-1}&{3}&{7}

$$\left| \begin{array} { r r r } { 5 } & { 6 } & { 3 } \\ { 0 } & { - 2 } & { 1 } \\ { - 1 } & { 3 } & { 7 } \end{array} \right|$$

Evaluate

$$-97$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&6&3\\0&-2&1\\-1&3&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&6&3&5&6\\0&-2&1&0&-2\\-1&3&7&-1&3\end{matrix}\right)$$

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

$$5\left(-2\right)\times 7+6\left(-1\right)=-76$$

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

$$-\left(-2\right)\times 3+3\times 5=21$$

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

$$-76-21$$

Subtract 21 from -76.

$$-97$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&6&3\\0&-2&1\\-1&3&7\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}-2&1\\3&7\end{matrix}\right))-6det(\left(\begin{matrix}0&1\\-1&7\end{matrix}\right))+3det(\left(\begin{matrix}0&-2\\-1&3\end{matrix}\right))$$

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

$$5\left(-2\times 7-3\right)-6\left(-\left(-1\right)\right)+3\left(-\left(-\left(-2\right)\right)\right)$$

Simplify.

$$5\left(-17\right)-6+3\left(-2\right)$$

Add the terms to obtain the final result.

$$-97$$

Factor

$$-97$$