Solve {rrr}{4}&{-1}&{-10}{11}&{5}&{-12}{1}&{2}&{2}

$$\left| \begin{array} { r r r } { 4 } & { - 1 } & { - 10 } \\ { 11 } & { 5 } & { - 12 } \\ { 1 } & { 2 } & { 2 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&-1&-10\\11&5&-12\\1&2&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&-1&-10&4&-1\\11&5&-12&11&5\\1&2&2&1&2\end{matrix}\right)$$

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

$$4\times 5\times 2-\left(-12\right)-10\times 11\times 2=-168$$

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

$$5\left(-10\right)+2\left(-12\right)\times 4+2\times 11\left(-1\right)=-168$$

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

$$-168-\left(-168\right)$$

Subtract -168 from -168.

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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).

$$det(\left(\begin{matrix}4&-1&-10\\11&5&-12\\1&2&2\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}5&-12\\2&2\end{matrix}\right))-\left(-det(\left(\begin{matrix}11&-12\\1&2\end{matrix}\right))\right)-10det(\left(\begin{matrix}11&5\\1&2\end{matrix}\right))$$

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

$$4\left(5\times 2-2\left(-12\right)\right)-\left(-\left(11\times 2-\left(-12\right)\right)\right)-10\left(11\times 2-5\right)$$

Simplify.

$$4\times 34-\left(-34\right)-10\times 17$$

Add the terms to obtain the final result.

$$0$$

Factor

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