Solve {ccc}{-2}&{-1}&{1}{-10}&{3}&{1}{2}&{0}&{1}

$$\left| \begin{array} { c c c } { - 2 } & { - 1 } & { 1 } \\ { - 10 } & { 3 } & { 1 } \\ { 2 } & { 0 } & { 1 } \end{array} \right|$$

Evaluate

$$-24$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&-1&1\\-10&3&1\\2&0&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&-1&1&-2&-1\\-10&3&1&-10&3\\2&0&1&2&0\end{matrix}\right)$$

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

$$-2\times 3-2=-8$$

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

$$2\times 3-10\left(-1\right)=16$$

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

$$-8-16$$

Subtract 16 from -8.

$$-24$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&-1&1\\-10&3&1\\2&0&1\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}3&1\\0&1\end{matrix}\right))-\left(-det(\left(\begin{matrix}-10&1\\2&1\end{matrix}\right))\right)+det(\left(\begin{matrix}-10&3\\2&0\end{matrix}\right))$$

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

$$-2\times 3-\left(-\left(-10-2\right)\right)-2\times 3$$

Simplify.

$$-2\times 3-\left(-\left(-12\right)\right)-6$$

Add the terms to obtain the final result.

$$-24$$

Factor

$$-24$$