Solve {lll}{1}&{1}&{1}{4}&{1}&{2}{0}&{1}&{6}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 4 } & { 1 } & { 2 } \\ { 0 } & { 1 } & { 6 } \end{array} \right|$$

Evaluate

$$-16$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\4&1&2\\0&1&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\4&1&2&4&1\\0&1&6&0&1\end{matrix}\right)$$

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

$$6+4=10$$

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

$$2+6\times 4=26$$

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

$$10-26$$

Subtract 26 from 10.

$$-16$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\4&1&2\\0&1&6\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&2\\1&6\end{matrix}\right))-det(\left(\begin{matrix}4&2\\0&6\end{matrix}\right))+det(\left(\begin{matrix}4&1\\0&1\end{matrix}\right))$$

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

$$6-2-4\times 6+4$$

Simplify.

$$4-24+4$$

Add the terms to obtain the final result.

$$-16$$

Factor

$$-16$$