Solve {ccc}{2}&{8}&{3}{10}&{7}&{2}{9}&{6}&{1}

$$\left| \begin{array} { c c c } { 2 } & { 8 } & { 3 } \\ { 10 } & { 7 } & { 2 } \\ { 9 } & { 6 } & { 1 } \end{array} \right|$$

Evaluate

$$45$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&8&3\\10&7&2\\9&6&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&8&3&2&8\\10&7&2&10&7\\9&6&1&9&6\end{matrix}\right)$$

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

$$2\times 7+8\times 2\times 9+3\times 10\times 6=338$$

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

$$9\times 7\times 3+6\times 2\times 2+10\times 8=293$$

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

$$338-293$$

Subtract 293 from 338.

$$45$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&8&3\\10&7&2\\9&6&1\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}7&2\\6&1\end{matrix}\right))-8det(\left(\begin{matrix}10&2\\9&1\end{matrix}\right))+3det(\left(\begin{matrix}10&7\\9&6\end{matrix}\right))$$

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

$$2\left(7-6\times 2\right)-8\left(10-9\times 2\right)+3\left(10\times 6-9\times 7\right)$$

Simplify.

$$2\left(-5\right)-8\left(-8\right)+3\left(-3\right)$$

Add the terms to obtain the final result.

$$45$$

Factor

$$3^{2}\times 5$$