Solve {rrr}{-2}&{1}&{3}{-2}&{1}&{3}{4}&{5}&{6}

$$\left| \begin{array} { r r r } { - 2 } & { 1 } & { 3 } \\ { - 2 } & { 1 } & { 3 } \\ { 4 } & { 5 } & { 6 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&1&3\\-2&1&3\\4&5&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&1&3&-2&1\\-2&1&3&-2&1\\4&5&6&4&5\end{matrix}\right)$$

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

$$-2\times 6+3\times 4+3\left(-2\right)\times 5=-30$$

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

$$4\times 3+5\times 3\left(-2\right)+6\left(-2\right)=-30$$

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

$$-30-\left(-30\right)$$

Subtract -30 from -30.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&1&3\\-2&1&3\\4&5&6\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}1&3\\5&6\end{matrix}\right))-det(\left(\begin{matrix}-2&3\\4&6\end{matrix}\right))+3det(\left(\begin{matrix}-2&1\\4&5\end{matrix}\right))$$

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

$$-2\left(6-5\times 3\right)-\left(-2\times 6-4\times 3\right)+3\left(-2\times 5-4\right)$$

Simplify.

$$-2\left(-9\right)-\left(-24\right)+3\left(-14\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$