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

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

Evaluate

$$56$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\left(-3\right)\left(-1\right)+5\times 5\times 3=84$$

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

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

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

$$84-28$$

Subtract 28 from 84.

$$56$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-3\right)-\left(-2\left(-5\right)\right)+5\times 15$$

Add the terms to obtain the final result.

$$56$$

Factor

$$2^{3}\times 7$$