Solve {lll}{5}&{7}&{8}{2}&{9}&{5}{8}&{7}&{6}

$$\left( \begin{array} { l l l } { 5 } & { 7 } & { 8 } \\ { 2 } & { 9 } & { 5 } \\ { 8 } & { 7 } & { 6 } \end{array} \right)$$

Calculate Determinant

$$-173$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&7&8\\2&9&5\\8&7&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&7&8&5&7\\2&9&5&2&9\\8&7&6&8&7\end{matrix}\right)$$

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

$$5\times 9\times 6+7\times 5\times 8+8\times 2\times 7=662$$

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

$$8\times 9\times 8+7\times 5\times 5+6\times 2\times 7=835$$

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

$$662-835$$

Subtract 835 from 662.

$$-173$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&7&8\\2&9&5\\8&7&6\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}9&5\\7&6\end{matrix}\right))-7det(\left(\begin{matrix}2&5\\8&6\end{matrix}\right))+8det(\left(\begin{matrix}2&9\\8&7\end{matrix}\right))$$

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

$$5\left(9\times 6-7\times 5\right)-7\left(2\times 6-8\times 5\right)+8\left(2\times 7-8\times 9\right)$$

Simplify.

$$5\times 19-7\left(-28\right)+8\left(-58\right)$$

Add the terms to obtain the final result.

$$-173$$

Evaluate

$$\left(\begin{matrix}5&7&8\\2&9&5\\8&7&6\end{matrix}\right)$$