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

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

Evaluate

$$-117$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-4\times 5\times 3+5\times 4\left(-3\right)=-120$$

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

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

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

$$-120-\left(-3\right)$$

Subtract -3 from -120.

$$-117$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-4\times 3-5\times 21$$

Add the terms to obtain the final result.

$$-117$$

Factor

$$-117$$