Solve {ccc}{1}&{3}&{-2}{2}&{-2}&{2}{-5}&{17}&{-14}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { - 2 } \\ { 2 } & { - 2 } & { 2 } \\ { - 5 } & { 17 } & { - 14 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&-2\\2&-2&2\\-5&17&-14\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&-2&1&3\\2&-2&2&2&-2\\-5&17&-14&-5&17\end{matrix}\right)$$

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

$$-2\left(-14\right)+3\times 2\left(-5\right)-2\times 2\times 17=-70$$

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

$$-5\left(-2\right)\left(-2\right)+17\times 2-14\times 2\times 3=-70$$

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

$$-70-\left(-70\right)$$

Subtract -70 from -70.

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Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&-2\\2&-2&2\\-5&17&-14\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-2&2\\17&-14\end{matrix}\right))-3det(\left(\begin{matrix}2&2\\-5&-14\end{matrix}\right))-2det(\left(\begin{matrix}2&-2\\-5&17\end{matrix}\right))$$

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

$$-2\left(-14\right)-17\times 2-3\left(2\left(-14\right)-\left(-5\times 2\right)\right)-2\left(2\times 17-\left(-5\left(-2\right)\right)\right)$$

Simplify.

$$-6-3\left(-18\right)-2\times 24$$

Add the terms to obtain the final result.

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Factor

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