Solve {lll}{8}&{9}&{7}{2}&{1}&{9}{1}&{2}&{0}

$$\left| \begin{array} { l l l } { 8 } & { 9 } & { 7 } \\ { 2 } & { 1 } & { 9 } \\ { 1 } & { 2 } & { 0 } \end{array} \right|$$

Evaluate

$$-42$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}8&9&7\\2&1&9\\1&2&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}8&9&7&8&9\\2&1&9&2&1\\1&2&0&1&2\end{matrix}\right)$$

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

$$9\times 9+7\times 2\times 2=109$$

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

$$7+2\times 9\times 8=151$$

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

$$109-151$$

Subtract 151 from 109.

$$-42$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}8&9&7\\2&1&9\\1&2&0\end{matrix}\right))$$

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.

$$8det(\left(\begin{matrix}1&9\\2&0\end{matrix}\right))-9det(\left(\begin{matrix}2&9\\1&0\end{matrix}\right))+7det(\left(\begin{matrix}2&1\\1&2\end{matrix}\right))$$

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

$$8\left(-2\times 9\right)-9\left(-9\right)+7\left(2\times 2-1\right)$$

Simplify.

$$8\left(-18\right)-9\left(-9\right)+7\times 3$$

Add the terms to obtain the final result.

$$-42$$

Factor

$$-42$$