Solve {ccc}{1}&{-2}&{2}{0}&{-6}&{11}{0}&{4}&{-4}

$$\left| \begin{array} { c c c } { 1 } & { - 2 } & { 2 } \\ { 0 } & { - 6 } & { 11 } \\ { 0 } & { 4 } & { - 4 } \end{array} \right|$$

Evaluate

$$-20$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-2&2\\0&-6&11\\0&4&-4\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-2&2&1&-2\\0&-6&11&0&-6\\0&4&-4&0&4\end{matrix}\right)$$

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

$$-6\left(-4\right)=24$$

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

$$4\times 11=44$$

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

$$24-44$$

Subtract 44 from 24.

$$-20$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-2&2\\0&-6&11\\0&4&-4\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-6&11\\4&-4\end{matrix}\right))-\left(-2det(\left(\begin{matrix}0&11\\0&-4\end{matrix}\right))\right)+2det(\left(\begin{matrix}0&-6\\0&4\end{matrix}\right))$$

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

$$-6\left(-4\right)-4\times 11$$

Simplify.

$$-20$$

Factor

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