Solve {lll}{0}&{3}&{2}{1}&{7}&{8}{0}&{5}&{4}

$$\left| \begin{array} { l l l } { 0 } & { 3 } & { 2 } \\ { 1 } & { 7 } & { 8 } \\ { 0 } & { 5 } & { 4 } \end{array} \right|$$

Evaluate

$$-2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&3&2\\1&7&8\\0&5&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&3&2&0&3\\1&7&8&1&7\\0&5&4&0&5\end{matrix}\right)$$

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

$$2\times 5=10$$

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

$$4\times 3=12$$

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

$$10-12$$

Subtract 12 from 10.

$$-2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&3&2\\1&7&8\\0&5&4\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}1&8\\0&4\end{matrix}\right))+2det(\left(\begin{matrix}1&7\\0&5\end{matrix}\right))$$

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

$$-3\times 4+2\times 5$$

Add the terms to obtain the final result.

$$-2$$

Factor

$$-2$$