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

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

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$6\times 4\times 4+7\times 5\times 2+8\times 3\times 3=238$$

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

$$2\times 4\times 8+3\times 5\times 6+4\times 3\times 7=238$$

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

$$238-238$$

Subtract 238 from 238.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$6-7\times 2+8$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$