Solve {ccc}{4}&{-3}&{0}{-6}&{4}&{2}{5}&{5}&{-1}

$$\left| \begin{array} { c c c } { 4 } & { - 3 } & { 0 } \\ { - 6 } & { 4 } & { 2 } \\ { 5 } & { 5 } & { - 1 } \end{array} \right|$$

Evaluate

$$-68$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&-3&0\\-6&4&2\\5&5&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&-3&0&4&-3\\-6&4&2&-6&4\\5&5&-1&5&5\end{matrix}\right)$$

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

$$4\times 4\left(-1\right)-3\times 2\times 5=-46$$

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

$$5\times 2\times 4-\left(-6\left(-3\right)\right)=22$$

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

$$-46-22$$

Subtract 22 from -46.

$$-68$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&-3&0\\-6&4&2\\5&5&-1\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}4&2\\5&-1\end{matrix}\right))-\left(-3det(\left(\begin{matrix}-6&2\\5&-1\end{matrix}\right))\right)$$

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

$$4\left(4\left(-1\right)-5\times 2\right)-\left(-3\left(-6\left(-1\right)-5\times 2\right)\right)$$

Simplify.

$$4\left(-14\right)-\left(-3\left(-4\right)\right)$$

Add the terms to obtain the final result.

$$-68$$

Factor

$$-68$$