Solve {lll}{3}&{4}&{1}{4}&{6}&{7}{7}&{8}&{9}

$$\left| \begin{array} { l l l } { 3 } & { 4 } & { 1 } \\ { 4 } & { 6 } & { 7 } \\ { 7 } & { 8 } & { 9 } \end{array} \right|$$

Evaluate

$$36$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&4&1\\4&6&7\\7&8&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&4&1&3&4\\4&6&7&4&6\\7&8&9&7&8\end{matrix}\right)$$

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

$$3\times 6\times 9+4\times 7\times 7+4\times 8=390$$

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

$$7\times 6+8\times 7\times 3+9\times 4\times 4=354$$

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

$$390-354$$

Subtract 354 from 390.

$$36$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&4&1\\4&6&7\\7&8&9\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}6&7\\8&9\end{matrix}\right))-4det(\left(\begin{matrix}4&7\\7&9\end{matrix}\right))+det(\left(\begin{matrix}4&6\\7&8\end{matrix}\right))$$

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

$$3\left(6\times 9-8\times 7\right)-4\left(4\times 9-7\times 7\right)+4\times 8-7\times 6$$

Simplify.

$$3\left(-2\right)-4\left(-13\right)-10$$

Add the terms to obtain the final result.

$$36$$

Factor

$$2^{2}\times 3^{2}$$