Solve {lll}{1}&{6}&{7}{2}&{5}&{8}{3}&{4}&{9}

$$\left( \begin{array} { l l l } { 1 } & { 6 } & { 7 } \\ { 2 } & { 5 } & { 8 } \\ { 3 } & { 4 } & { 9 } \end{array} \right)$$

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&6&7\\2&5&8\\3&4&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&6&7&1&6\\2&5&8&2&5\\3&4&9&3&4\end{matrix}\right)$$

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

$$5\times 9+6\times 8\times 3+7\times 2\times 4=245$$

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

$$3\times 5\times 7+4\times 8+9\times 2\times 6=245$$

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

$$245-245$$

Subtract 245 from 245.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&6&7\\2&5&8\\3&4&9\end{matrix}\right))$$

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.

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

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

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

Simplify.

$$13-6\left(-6\right)+7\left(-7\right)$$

Add the terms to obtain the final result.

$$0$$

Evaluate

$$\left(\begin{matrix}1&6&7\\2&5&8\\3&4&9\end{matrix}\right)$$