Solve {rrr}{1}&{3}&{0}{-1}&{2}&{4}{8}&{5}&{3}=

$$\left| \begin{array} { r r r } { 1 } & { 3 } & { 0 } \\ { - 1 } & { 2 } & { 4 } \\ { 8 } & { 5 } & { 3 } \end{array} \right| =$$

Evaluate

$$91$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&0\\-1&2&4\\8&5&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&0&1&3\\-1&2&4&-1&2\\8&5&3&8&5\end{matrix}\right)$$

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

$$2\times 3+3\times 4\times 8=102$$

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

$$5\times 4+3\left(-1\right)\times 3=11$$

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

$$102-11$$

Subtract 11 from 102.

$$91$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&0\\-1&2&4\\8&5&3\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&4\\5&3\end{matrix}\right))-3det(\left(\begin{matrix}-1&4\\8&3\end{matrix}\right))$$

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

$$2\times 3-5\times 4-3\left(-3-8\times 4\right)$$

Simplify.

$$-14-3\left(-35\right)$$

Add the terms to obtain the final result.

$$91$$

Factor

$$7\times 13$$