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

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

Evaluate

$$-78$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 2\left(-3\right)+3\left(-3\right)-5\left(-1\right)\times 3=-24$$

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

$$-3\times 2\left(-5\right)+3\times 5-3\left(-1\right)\times 3=54$$

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

$$-24-54$$

Subtract 54 from -24.

$$-78$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5\left(-9\right)-3\times 6-5\times 3$$

Add the terms to obtain the final result.

$$-78$$

Factor

$$-78$$