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

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

Evaluate

$$-131$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$7\times 3+2\times 6\times 5+8\times 2\times 8=209$$

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

$$5\times 7\times 8+8\times 6+3\times 2\times 2=340$$

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

$$209-340$$

Subtract 340 from 209.

$$-131$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-27-2\left(-24\right)+8\left(-19\right)$$

Add the terms to obtain the final result.

$$-131$$

Factor

$$-131$$