Solve {ccc}{12}&{9}&{3}{37}&{8}&{4}{89}&{7}&{6}

$$\left| \begin{array} { c c c } { 12 } & { 9 } & { 3 } \\ { 37 } & { 8 } & { 4 } \\ { 89 } & { 7 } & { 6 } \end{array} \right|$$

Evaluate

$$87$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}12&9&3\\37&8&4\\89&7&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}12&9&3&12&9\\37&8&4&37&8\\89&7&6&89&7\end{matrix}\right)$$

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

$$12\times 8\times 6+9\times 4\times 89+3\times 37\times 7=4557$$

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

$$89\times 8\times 3+7\times 4\times 12+6\times 37\times 9=4470$$

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

$$4557-4470$$

Subtract 4470 from 4557.

$$87$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}12&9&3\\37&8&4\\89&7&6\end{matrix}\right))$$

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.

$$12det(\left(\begin{matrix}8&4\\7&6\end{matrix}\right))-9det(\left(\begin{matrix}37&4\\89&6\end{matrix}\right))+3det(\left(\begin{matrix}37&8\\89&7\end{matrix}\right))$$

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

$$12\left(8\times 6-7\times 4\right)-9\left(37\times 6-89\times 4\right)+3\left(37\times 7-89\times 8\right)$$

Simplify.

$$12\times 20-9\left(-134\right)+3\left(-453\right)$$

Add the terms to obtain the final result.

$$87$$

Factor

$$3\times 29$$