Solve {lll}{1}&{1}&{1}{1}&{0}&{-1}{2}&{0}&{1}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { 0 } & { - 1 } \\ { 2 } & { 0 } & { 1 } \end{array} \right|$$

Evaluate

$$-3$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2=-2$$

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

$$1=1$$

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

$$-2-1$$

Subtract 1 from -2.

$$-3$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-\left(1-2\left(-1\right)\right)$$

Simplify.

$$-3$$

Factor

$$-3$$