Solve {rrr}{1}&{-1}&{2}{1}&{3}&{-4}{-5}&{3}&{-3}

$$\left| \begin{array} { r r r } { 1 } & { - 1 } & { 2 } \\ { 1 } & { 3 } & { - 4 } \\ { - 5 } & { 3 } & { - 3 } \end{array} \right|$$

Evaluate

$$16$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-1&2\\1&3&-4\\-5&3&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-1&2&1&-1\\1&3&-4&1&3\\-5&3&-3&-5&3\end{matrix}\right)$$

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

$$3\left(-3\right)-\left(-4\left(-5\right)\right)+2\times 3=-23$$

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

$$-5\times 3\times 2+3\left(-4\right)-3\left(-1\right)=-39$$

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

$$-23-\left(-39\right)$$

Subtract -39 from -23.

$$16$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-1&2\\1&3&-4\\-5&3&-3\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}3&-4\\3&-3\end{matrix}\right))-\left(-det(\left(\begin{matrix}1&-4\\-5&-3\end{matrix}\right))\right)+2det(\left(\begin{matrix}1&3\\-5&3\end{matrix}\right))$$

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

$$3\left(-3\right)-3\left(-4\right)-\left(-\left(-3-\left(-5\left(-4\right)\right)\right)\right)+2\left(3-\left(-5\times 3\right)\right)$$

Simplify.

$$3-\left(-\left(-23\right)\right)+2\times 18$$

Add the terms to obtain the final result.

$$16$$

Factor

$$2^{4}$$