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

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 4 } \\ { 0 } & { 1 } & { 1 } \\ { - 1 } & { 2 } & { 0 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-4+2=-2$$

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

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

Subtract -2 from -2.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2-2+4$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$