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

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$5\times 4\left(-1\right)-4\left(-2\right)\times 3+3\times 2=10$$

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

$$4-10$$

Subtract 10 from 4.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-4\right)-2\times 13-\left(-32\right)$$

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$