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

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

Evaluate

$$-145$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\times 5-4\times 3\times 3-5\times 4=-66$$

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

$$3+4\times 3\left(-2\right)+5\left(-5\right)\left(-4\right)=79$$

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

$$-66-79$$

Subtract 79 from -66.

$$-145$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2\left(-7\right)-\left(-4\left(-34\right)\right)-23$$

Add the terms to obtain the final result.

$$-145$$

Factor

$$-145$$