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

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\times 2\left(-1\right)+3\times 4=16$$

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

$$4\times 2\times 4+5\left(-2\right)=22$$

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

$$16-22$$

Subtract 22 from 16.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$4\left(-8\right)-\left(-2\times 7\right)+3\times 4$$

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$