Solve {ccc}{3}&{-2}&{5}{0}&{3}&{-4}{0}&{2}&{-2}

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

Evaluate

$$6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\left(-2\right)=-18$$

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

$$2\left(-4\right)\times 3=-24$$

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

$$-18-\left(-24\right)$$

Subtract -24 from -18.

$$6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 2$$

Add the terms to obtain the final result.

$$6$$

Factor

$$2\times 3$$