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

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

Calculate Determinant

$$6$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-1+2\left(-1\right)=-3$$

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

$$3-\left(-3\right)$$

Subtract -3 from 3.

$$6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$1-\left(-4\right)+1$$

Add the terms to obtain the final result.

$$6$$

Evaluate

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