Solve {rrr}{1}&{1}&{2}{3}&{-1}&{2}{5}&{1}&{6}

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$-6+2\times 5+2\times 3=10$$

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

$$5\left(-1\right)\times 2+2+6\times 3=10$$

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

$$10-10$$

Subtract 10 from 10.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-6-2-\left(3\times 6-5\times 2\right)+2\left(3-5\left(-1\right)\right)$$

Simplify.

$$-8-8+2\times 8$$

Add the terms to obtain the final result.

$$0$$

Factor

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