Solve {ccc}{1}&{1}&{-9}{11}&{-4}&{1}{-7}&{1}&{1}

$$\left( \begin{array} { c c c } { 1 } & { 1 } & { - 9 } \\ { 11 } & { - 4 } & { 1 } \\ { - 7 } & { 1 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$130$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&-9\\11&-4&1\\-7&1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&-9&1&1\\11&-4&1&11&-4\\-7&1&1&-7&1\end{matrix}\right)$$

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

$$-4-7-9\times 11=-110$$

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

$$-7\left(-4\right)\left(-9\right)+1+11=-240$$

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

$$-110-\left(-240\right)$$

Subtract -240 from -110.

$$130$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&-9\\11&-4&1\\-7&1&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-4&1\\1&1\end{matrix}\right))-det(\left(\begin{matrix}11&1\\-7&1\end{matrix}\right))-9det(\left(\begin{matrix}11&-4\\-7&1\end{matrix}\right))$$

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

$$-4-1-\left(11-\left(-7\right)\right)-9\left(11-\left(-7\left(-4\right)\right)\right)$$

Simplify.

$$-5-18-9\left(-17\right)$$

Add the terms to obtain the final result.

$$130$$

Evaluate

$$\left(\begin{matrix}1&1&-9\\11&-4&1\\-7&1&1\end{matrix}\right)$$