Solve {lll}{5}&{6}&{7}{8}&{0}&{9}{4}&{3}&{2}

$$\left| \begin{array} { l l l } { 5 } & { 6 } & { 7 } \\ { 8 } & { 0 } & { 9 } \\ { 4 } & { 3 } & { 2 } \end{array} \right|$$

Evaluate

$$153$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&6&7\\8&0&9\\4&3&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&6&7&5&6\\8&0&9&8&0\\4&3&2&4&3\end{matrix}\right)$$

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

$$6\times 9\times 4+7\times 8\times 3=384$$

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

$$3\times 9\times 5+2\times 8\times 6=231$$

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

$$384-231$$

Subtract 231 from 384.

$$153$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&6&7\\8&0&9\\4&3&2\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}0&9\\3&2\end{matrix}\right))-6det(\left(\begin{matrix}8&9\\4&2\end{matrix}\right))+7det(\left(\begin{matrix}8&0\\4&3\end{matrix}\right))$$

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

$$5\left(-3\times 9\right)-6\left(8\times 2-4\times 9\right)+7\times 8\times 3$$

Simplify.

$$5\left(-27\right)-6\left(-20\right)+7\times 24$$

Add the terms to obtain the final result.

$$153$$

Factor

$$3^{2}\times 17$$