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

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

Evaluate

$$-54$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\left(-4\right)-2\left(-2\right)\times 2+4\left(-2\right)=-12$$

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

$$2\times 3-2\left(-2\right)-4\times 4\left(-2\right)=42$$

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

$$-12-42$$

Subtract 42 from -12.

$$-54$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-16-\left(-2\left(-12\right)\right)-14$$

Add the terms to obtain the final result.

$$-54$$

Factor

$$-54$$