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

$$\left( \begin{array} { c c c } { 0 } & { 1 } & { 2 } \\ { 1 } & { 1 } & { 4 } \\ { 2 } & { - 1 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$2$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 2+2\left(-1\right)=6$$

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

$$2\times 2=4$$

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

$$6-4$$

Subtract 4 from 6.

$$2$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-\left(-2\times 4\right)+2\left(-1-2\right)$$

Simplify.

$$-\left(-8\right)+2\left(-3\right)$$

Add the terms to obtain the final result.

$$2$$

Evaluate

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