Solve {ccc}{1}&{1}&{1}{-3}&{0}&{-6}{2}&{1}&{3}

$$\left| \begin{array} { c c c } { 1 } & { 1 } & { 1 } \\ { - 3 } & { 0 } & { - 6 } \\ { 2 } & { 1 } & { 3 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\-3&0&-6\\2&1&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\-3&0&-6&-3&0\\2&1&3&2&1\end{matrix}\right)$$

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

$$-6\times 2-3=-15$$

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

$$-6+3\left(-3\right)=-15$$

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

$$-15-\left(-15\right)$$

Subtract -15 from -15.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\-3&0&-6\\2&1&3\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}0&-6\\1&3\end{matrix}\right))-det(\left(\begin{matrix}-3&-6\\2&3\end{matrix}\right))+det(\left(\begin{matrix}-3&0\\2&1\end{matrix}\right))$$

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

$$-\left(-6\right)-\left(-3\times 3-2\left(-6\right)\right)-3$$

Simplify.

$$6-3-3$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$