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

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

Calculate Determinant

$$35$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\left(-1\right)+3\times 3\times 5=47$$

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

$$-4\times 3+8\times 3=12$$

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

$$47-12$$

Subtract 12 from 47.

$$35$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-22+3\times 19$$

Add the terms to obtain the final result.

$$35$$

Evaluate

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