Solve {ccc}{9}&{9}&{12}{1}&{-3}&{-4}{1}&{9}&{12}

$$\left| \begin{array} { c c c } { 9 } & { 9 } & { 12 } \\ { 1 } & { - 3 } & { - 4 } \\ { 1 } & { 9 } & { 12 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}9&9&12\\1&-3&-4\\1&9&12\end{matrix}\right))$$

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

$$\left(\begin{matrix}9&9&12&9&9\\1&-3&-4&1&-3\\1&9&12&1&9\end{matrix}\right)$$

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

$$9\left(-3\right)\times 12+9\left(-4\right)+12\times 9=-252$$

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

$$-3\times 12+9\left(-4\right)\times 9+12\times 9=-252$$

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

$$-252-\left(-252\right)$$

Subtract -252 from -252.

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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).

$$det(\left(\begin{matrix}9&9&12\\1&-3&-4\\1&9&12\end{matrix}\right))$$

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.

$$9det(\left(\begin{matrix}-3&-4\\9&12\end{matrix}\right))-9det(\left(\begin{matrix}1&-4\\1&12\end{matrix}\right))+12det(\left(\begin{matrix}1&-3\\1&9\end{matrix}\right))$$

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

$$9\left(-3\times 12-9\left(-4\right)\right)-9\left(12-\left(-4\right)\right)+12\left(9-\left(-3\right)\right)$$

Simplify.

$$-9\times 16+12\times 12$$

Add the terms to obtain the final result.

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Factor

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