Solve {lll}{3}&{2}&{1}{4}&{2}&{3}{7}&{5}&{9}

$$\left| \begin{array} { l l l } { 3 } & { 2 } & { 1 } \\ { 4 } & { 2 } & { 3 } \\ { 7 } & { 5 } & { 9 } \end{array} \right|$$

Evaluate

$$-15$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&2&1\\4&2&3\\7&5&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&2&1&3&2\\4&2&3&4&2\\7&5&9&7&5\end{matrix}\right)$$

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

$$3\times 2\times 9+2\times 3\times 7+4\times 5=116$$

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

$$7\times 2+5\times 3\times 3+9\times 4\times 2=131$$

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

$$116-131$$

Subtract 131 from 116.

$$-15$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&2&1\\4&2&3\\7&5&9\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}2&3\\5&9\end{matrix}\right))-2det(\left(\begin{matrix}4&3\\7&9\end{matrix}\right))+det(\left(\begin{matrix}4&2\\7&5\end{matrix}\right))$$

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

$$3\left(2\times 9-5\times 3\right)-2\left(4\times 9-7\times 3\right)+4\times 5-7\times 2$$

Simplify.

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

Add the terms to obtain the final result.

$$-15$$

Factor

$$-15$$