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

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

Evaluate

$$61$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 2+3\left(-2\right)\left(-3\right)=28$$

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

$$3\left(-2\right)\times 5-3=-33$$

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

$$28-\left(-33\right)$$

Subtract -33 from 28.

$$61$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$5\left(2-3\left(-2\right)\right)-3\left(-1-\left(-3\left(-2\right)\right)\right)$$

Simplify.

$$5\times 8-3\left(-7\right)$$

Add the terms to obtain the final result.

$$61$$

Factor

$$61$$