Solve {lll}{1}&{2}&{4}{0}&{9}&{8}{3}&{2}&{1}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 4 } \\ { 0 } & { 9 } & { 8 } \\ { 3 } & { 2 } & { 1 } \end{array} \right|$$

Evaluate

$$-67$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&4\\0&9&8\\3&2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&4&1&2\\0&9&8&0&9\\3&2&1&3&2\end{matrix}\right)$$

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

$$9+2\times 8\times 3=57$$

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

$$3\times 9\times 4+2\times 8=124$$

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

$$57-124$$

Subtract 124 from 57.

$$-67$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&4\\0&9&8\\3&2&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}9&8\\2&1\end{matrix}\right))-2det(\left(\begin{matrix}0&8\\3&1\end{matrix}\right))+4det(\left(\begin{matrix}0&9\\3&2\end{matrix}\right))$$

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

$$9-2\times 8-2\left(-3\times 8\right)+4\left(-3\times 9\right)$$

Simplify.

$$-7-2\left(-24\right)+4\left(-27\right)$$

Add the terms to obtain the final result.

$$-67$$

Factor

$$-67$$