Solve {lll}{1}&{3}&{6}{7}&{4}&{9}{1}&{1}&{1}

$$\left| \begin{array} { l l l } { 1 } & { 3 } & { 6 } \\ { 7 } & { 4 } & { 9 } \\ { 1 } & { 1 } & { 1 } \end{array} \right|$$

Evaluate

$$19$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&6\\7&4&9\\1&1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&6&1&3\\7&4&9&7&4\\1&1&1&1&1\end{matrix}\right)$$

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

$$4+3\times 9+6\times 7=73$$

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

$$4\times 6+9+7\times 3=54$$

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

$$73-54$$

Subtract 54 from 73.

$$19$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&6\\7&4&9\\1&1&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}4&9\\1&1\end{matrix}\right))-3det(\left(\begin{matrix}7&9\\1&1\end{matrix}\right))+6det(\left(\begin{matrix}7&4\\1&1\end{matrix}\right))$$

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

$$4-9-3\left(7-9\right)+6\left(7-4\right)$$

Simplify.

$$-5-3\left(-2\right)+6\times 3$$

Add the terms to obtain the final result.

$$19$$

Factor

$$19$$