Solve {lll}{2}&{3}&{6}{5}&{8}&{1}{6}&{9}&{5}

$$\left( \begin{array} { l l l } { 2 } & { 3 } & { 6 } \\ { 5 } & { 8 } & { 1 } \\ { 6 } & { 9 } & { 5 } \end{array} \right)$$

Calculate Determinant

$$-13$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&6\\5&8&1\\6&9&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&6&2&3\\5&8&1&5&8\\6&9&5&6&9\end{matrix}\right)$$

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

$$2\times 8\times 5+3\times 6+6\times 5\times 9=368$$

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

$$6\times 8\times 6+9\times 2+5\times 5\times 3=381$$

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

$$368-381$$

Subtract 381 from 368.

$$-13$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&6\\5&8&1\\6&9&5\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}8&1\\9&5\end{matrix}\right))-3det(\left(\begin{matrix}5&1\\6&5\end{matrix}\right))+6det(\left(\begin{matrix}5&8\\6&9\end{matrix}\right))$$

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

$$2\left(8\times 5-9\right)-3\left(5\times 5-6\right)+6\left(5\times 9-6\times 8\right)$$

Simplify.

$$2\times 31-3\times 19+6\left(-3\right)$$

Add the terms to obtain the final result.

$$-13$$

Evaluate

$$\left(\begin{matrix}2&3&6\\5&8&1\\6&9&5\end{matrix}\right)$$