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

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

Evaluate

$$-5$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-5\left(-2\right)\left(-3\right)+3\times 2-4\times 3=-36$$

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

$$-2\left(-4\right)+3\times 2\left(-5\right)-3\times 3=-31$$

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

$$-36-\left(-31\right)$$

Subtract -31 from -36.

$$-5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-5$$

Factor

$$-5$$