Solve {rrr}{4}&{-1}&{4}{2}&{1}&{8}{1}&{2}&{1}

$$\left| \begin{array} { r r r } { 4 } & { - 1 } & { 4 } \\ { 2 } & { 1 } & { 8 } \\ { 1 } & { 2 } & { 1 } \end{array} \right|$$

Evaluate

$$-54$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&-1&4\\2&1&8\\1&2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&-1&4&4&-1\\2&1&8&2&1\\1&2&1&1&2\end{matrix}\right)$$

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

$$4-8+4\times 2\times 2=12$$

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

$$4+2\times 8\times 4+2\left(-1\right)=66$$

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

$$12-66$$

Subtract 66 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).

$$det(\left(\begin{matrix}4&-1&4\\2&1&8\\1&2&1\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}1&8\\2&1\end{matrix}\right))-\left(-det(\left(\begin{matrix}2&8\\1&1\end{matrix}\right))\right)+4det(\left(\begin{matrix}2&1\\1&2\end{matrix}\right))$$

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

$$4\left(1-2\times 8\right)-\left(-\left(2-8\right)\right)+4\left(2\times 2-1\right)$$

Simplify.

$$4\left(-15\right)-\left(-\left(-6\right)\right)+4\times 3$$

Add the terms to obtain the final result.

$$-54$$

Factor

$$-54$$