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

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 8 } \\ { 2 } & { 7 } & { 6 } \\ { 3 } & { 7 } & { 4 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$7\times 4+2\times 6\times 3+8\times 2\times 7=176$$

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

$$3\times 7\times 8+7\times 6+4\times 2\times 2=226$$

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

$$176-226$$

Subtract 226 from 176.

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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).

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

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.

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

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

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

Simplify.

$$-14-2\left(-10\right)+8\left(-7\right)$$

Add the terms to obtain the final result.

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Factor

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