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

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 7+2\times 2\times 5+3\times 2\times 6=70$$

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

$$5\times 2\times 3+6\times 2+7\times 2\times 2=70$$

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

$$70-70$$

Subtract 70 from 70.

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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).

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

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.

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

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

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

Simplify.

$$2-2\times 4+3\times 2$$

Add the terms to obtain the final result.

$$0$$

Factor

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