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

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

Evaluate

$$-17$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\left(-3\right)\left(-5\right)+4\times 4\left(-2\right)+2\times 3=19$$

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

$$2-19$$

Subtract 19 from 2.

$$-17$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2\left(-19\right)-3\left(-10\right)-5\times 17$$

Add the terms to obtain the final result.

$$-17$$

Factor

$$-17$$