Solve {lll}{1}&{2}&{4}{1}&{4}&{1}{6}&{8}&{5}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 4 } \\ { 1 } & { 4 } & { 1 } \\ { 6 } & { 8 } & { 5 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 5+2\times 6+4\times 8=64$$

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

$$6\times 4\times 4+8+5\times 2=114$$

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

$$64-114$$

Subtract 114 from 64.

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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).

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

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.

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

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

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

Simplify.

$$12-2\left(-1\right)+4\left(-16\right)$$

Add the terms to obtain the final result.

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Factor

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