Solve {lll}{1}&{0}&{2}{3}&{5}&{2}{6}&{6}&{7}

$$\left| \begin{array} { l l l } { 1 } & { 0 } & { 2 } \\ { 3 } & { 5 } & { 2 } \\ { 6 } & { 6 } & { 7 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&2\\3&5&2\\6&6&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&2&1&0\\3&5&2&3&5\\6&6&7&6&6\end{matrix}\right)$$

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

$$5\times 7+2\times 3\times 6=71$$

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

$$6\times 5\times 2+6\times 2=72$$

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

$$71-72$$

Subtract 72 from 71.

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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).

$$det(\left(\begin{matrix}1&0&2\\3&5&2\\6&6&7\end{matrix}\right))$$

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.

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

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

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

Simplify.

$$23+2\left(-12\right)$$

Add the terms to obtain the final result.

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Factor

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