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

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

Evaluate

$$-15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\times 8+5\times 6+2\times 4\times 7=158$$

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

$$6\times 3\times 2+7\times 5\times 3+8\times 4=173$$

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

$$158-173$$

Subtract 173 from 158.

$$-15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-11\right)-2+2\times 10$$

Add the terms to obtain the final result.

$$-15$$

Factor

$$-15$$