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

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

Evaluate

$$-7$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

$$3-10$$

Subtract 10 from 3.

$$-7$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-13\right)-3-\left(-22\right)$$

Add the terms to obtain the final result.

$$-7$$

Factor

$$-7$$