Solve {ccc}{1}&{3}&{-2}{-3}&{-9}&{6}{11}&{7}&{5}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { - 2 } \\ { - 3 } & { - 9 } & { 6 } \\ { 11 } & { 7 } & { 5 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&-2\\-3&-9&6\\11&7&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&-2&1&3\\-3&-9&6&-3&-9\\11&7&5&11&7\end{matrix}\right)$$

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

$$-9\times 5+3\times 6\times 11-2\left(-3\right)\times 7=195$$

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

$$11\left(-9\right)\left(-2\right)+7\times 6+5\left(-3\right)\times 3=195$$

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

$$195-195$$

Subtract 195 from 195.

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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).

$$det(\left(\begin{matrix}1&3&-2\\-3&-9&6\\11&7&5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-9&6\\7&5\end{matrix}\right))-3det(\left(\begin{matrix}-3&6\\11&5\end{matrix}\right))-2det(\left(\begin{matrix}-3&-9\\11&7\end{matrix}\right))$$

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

$$-9\times 5-7\times 6-3\left(-3\times 5-11\times 6\right)-2\left(-3\times 7-11\left(-9\right)\right)$$

Simplify.

$$-87-3\left(-81\right)-2\times 78$$

Add the terms to obtain the final result.

$$0$$

Factor

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