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

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

Calculate Determinant

$$15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 2+3\times 2\times 4=32$$

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

$$-3+2\times 2\times 5=17$$

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

$$32-17$$

Subtract 17 from 32.

$$15$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$4\times 2-5\times 2\times 2+3\left(2\times 4-\left(-1\right)\right)$$

Simplify.

$$4\times 2-5\times 4+3\times 9$$

Add the terms to obtain the final result.

$$15$$

Evaluate

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