Solve {lll}{1}&{7}&{9}{0}&{2}&{6}{0}&{8}&{7}

$$\left| \begin{array} { l l l } { 1 } & { 7 } & { 9 } \\ { 0 } & { 2 } & { 6 } \\ { 0 } & { 8 } & { 7 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&7&9\\0&2&6\\0&8&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&7&9&1&7\\0&2&6&0&2\\0&8&7&0&8\end{matrix}\right)$$

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

$$2\times 7=14$$

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

$$8\times 6=48$$

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

$$14-48$$

Subtract 48 from 14.

$$-34$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&7&9\\0&2&6\\0&8&7\end{matrix}\right))$$

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.

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

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

$$2\times 7-8\times 6$$

Simplify.

$$-34$$

Factor

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