Solve {ccc}{3}&{5}&{9}{4}&{2}&{2}{4}&{3}&{2}

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

Evaluate

$$30$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 2\times 2+5\times 2\times 4+9\times 4\times 3=160$$

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

$$4\times 2\times 9+3\times 2\times 3+2\times 4\times 5=130$$

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

$$160-130$$

Subtract 130 from 160.

$$30$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$30$$

Factor

$$2\times 3\times 5$$