Solve {ccc}{3}&{-2}&{1}{5}&{3}&{0}{1}&{1}&{-2}

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

Evaluate

$$-36$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\left(-2\right)+5=-13$$

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

$$3-2\times 5\left(-2\right)=23$$

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

$$-13-23$$

Subtract 23 from -13.

$$-36$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-6\right)-\left(-2\left(-10\right)\right)+2$$

Add the terms to obtain the final result.

$$-36$$

Factor

$$-36$$