Solve {lll}{3}&{7}&{2}{2}&{6}&{7}{1}&{5}&{3}

$$\left| \begin{array} { l l l } { 3 } & { 7 } & { 2 } \\ { 2 } & { 6 } & { 7 } \\ { 1 } & { 5 } & { 3 } \end{array} \right|$$

Evaluate

$$-36$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&7&2\\2&6&7\\1&5&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&7&2&3&7\\2&6&7&2&6\\1&5&3&1&5\end{matrix}\right)$$

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

$$3\times 6\times 3+7\times 7+2\times 2\times 5=123$$

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

$$6\times 2+5\times 7\times 3+3\times 2\times 7=159$$

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

$$123-159$$

Subtract 159 from 123.

$$-36$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&7&2\\2&6&7\\1&5&3\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}6&7\\5&3\end{matrix}\right))-7det(\left(\begin{matrix}2&7\\1&3\end{matrix}\right))+2det(\left(\begin{matrix}2&6\\1&5\end{matrix}\right))$$

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

$$3\left(6\times 3-5\times 7\right)-7\left(2\times 3-7\right)+2\left(2\times 5-6\right)$$

Simplify.

$$3\left(-17\right)-7\left(-1\right)+2\times 4$$

Add the terms to obtain the final result.

$$-36$$

Factor

$$-36$$