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

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

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$\text{true}$$

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

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$0$$

Factor

$$0$$