Solve {ccc}{-3}&{0}&{5}{3}&{4}&{-4}{-5}&{4}&{-6}

$$\left| \begin{array} { c c c } { - 3 } & { 0 } & { 5 } \\ { 3 } & { 4 } & { - 4 } \\ { - 5 } & { 4 } & { - 6 } \end{array} \right|$$

Evaluate

$$184$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-3&0&5\\3&4&-4\\-5&4&-6\end{matrix}\right))$$

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

$$\left(\begin{matrix}-3&0&5&-3&0\\3&4&-4&3&4\\-5&4&-6&-5&4\end{matrix}\right)$$

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

$$-3\times 4\left(-6\right)+5\times 3\times 4=132$$

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

$$-5\times 4\times 5+4\left(-4\right)\left(-3\right)=-52$$

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

$$132-\left(-52\right)$$

Subtract -52 from 132.

$$184$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-3&0&5\\3&4&-4\\-5&4&-6\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}4&-4\\4&-6\end{matrix}\right))+5det(\left(\begin{matrix}3&4\\-5&4\end{matrix}\right))$$

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

$$-3\left(4\left(-6\right)-4\left(-4\right)\right)+5\left(3\times 4-\left(-5\times 4\right)\right)$$

Simplify.

$$-3\left(-8\right)+5\times 32$$

Add the terms to obtain the final result.

$$184$$

Factor

$$2^{3}\times 23$$