Solve {ccc}{1}&{1}&{2}{-3}&{4}&{-6}{-2}&{3}&{-2}

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

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\left(-2\right)-6\left(-2\right)+2\left(-3\right)\times 3=-14$$

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

$$-2\times 4\times 2+3\left(-6\right)-2\left(-3\right)=-28$$

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

$$-14-\left(-28\right)$$

Subtract -28 from -14.

$$14$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$10-\left(-6\right)+2\left(-1\right)$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$