Solve {lll}{1}&{0}&{2}{3}&{4}&{5}{6}&{7}&{8}

$$\left| \begin{array} { l l l } { 1 } & { 0 } & { 2 } \\ { 3 } & { 4 } & { 5 } \\ { 6 } & { 7 } & { 8 } \end{array} \right|$$

Evaluate

$$-9$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&2\\3&4&5\\6&7&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&2&1&0\\3&4&5&3&4\\6&7&8&6&7\end{matrix}\right)$$

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

$$4\times 8+2\times 3\times 7=74$$

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

$$6\times 4\times 2+7\times 5=83$$

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

$$74-83$$

Subtract 83 from 74.

$$-9$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&2\\3&4&5\\6&7&8\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&5\\7&8\end{matrix}\right))+2det(\left(\begin{matrix}3&4\\6&7\end{matrix}\right))$$

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

$$4\times 8-7\times 5+2\left(3\times 7-6\times 4\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$-9$$

Factor

$$-9$$