Solve {ccc}{1}&{6}&{-3}{2}&{7}&{5}{1}&{1}&{8}

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

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$7\times 8+6\times 5-3\times 2=80$$

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

$$7\left(-3\right)+5+8\times 2\times 6=80$$

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

$$80-80$$

Subtract 80 from 80.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$51-6\times 11-3\left(-5\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$