Solve {rrr}{-2}&{-4}&{1}{3}&{0}&{3}{5}&{4}&{-2}

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

Evaluate

$$-48$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-4\times 3\times 5+3\times 4=-48$$

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

$$4\times 3\left(-2\right)-2\times 3\left(-4\right)=0$$

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

$$-48$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2\left(-12\right)-\left(-4\left(-21\right)\right)+12$$

Add the terms to obtain the final result.

$$-48$$

Factor

$$-48$$