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

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

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$9\times 5+8\times 4\times 3+7\times 6\times 2=225$$

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

$$3\times 5\times 7+2\times 4\times 9+6\times 8=225$$

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

$$225-225$$

Subtract 225 from 225.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$9\left(-3\right)-8\left(-6\right)+7\left(-3\right)$$

Add the terms to obtain the final result.

$$0$$

Evaluate

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