Solve {lll}{5}&{4}&{1}{0}&{9}&{7}{8}&{1}&{0}

$$\left| \begin{array} { l l l } { 5 } & { 4 } & { 1 } \\ { 0 } & { 9 } & { 7 } \\ { 8 } & { 1 } & { 0 } \end{array} \right|$$

Evaluate

$$117$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&4&1\\0&9&7\\8&1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&4&1&5&4\\0&9&7&0&9\\8&1&0&8&1\end{matrix}\right)$$

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

$$4\times 7\times 8=224$$

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

$$8\times 9+7\times 5=107$$

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

$$224-107$$

Subtract 107 from 224.

$$117$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&4&1\\0&9&7\\8&1&0\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}9&7\\1&0\end{matrix}\right))-4det(\left(\begin{matrix}0&7\\8&0\end{matrix}\right))+det(\left(\begin{matrix}0&9\\8&1\end{matrix}\right))$$

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

$$5\left(-7\right)-4\left(-8\times 7\right)-8\times 9$$

Simplify.

$$5\left(-7\right)-4\left(-56\right)-72$$

Add the terms to obtain the final result.

$$117$$

Factor

$$3^{2}\times 13$$