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

$$\left| \begin{array} { l l l } { 9 } & { 1 } & { 7 } \\ { 6 } & { 2 } & { 5 } \\ { 3 } & { 2 } & { 1 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$9\times 2+5\times 3+7\times 6\times 2=117$$

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

$$3\times 2\times 7+2\times 5\times 9+6=138$$

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

$$117-138$$

Subtract 138 from 117.

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Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

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Factor

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