Solve {lll}{1}&{2}&{4}{4}&{3}&{1}{8}&{7}&{5}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 4 } \\ { 4 } & { 3 } & { 1 } \\ { 8 } & { 7 } & { 5 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&4\\4&3&1\\8&7&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&4&1&2\\4&3&1&4&3\\8&7&5&8&7\end{matrix}\right)$$

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

$$3\times 5+2\times 8+4\times 4\times 7=143$$

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

$$8\times 3\times 4+7+5\times 4\times 2=143$$

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

$$143-143$$

Subtract 143 from 143.

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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).

$$det(\left(\begin{matrix}1&2&4\\4&3&1\\8&7&5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}3&1\\7&5\end{matrix}\right))-2det(\left(\begin{matrix}4&1\\8&5\end{matrix}\right))+4det(\left(\begin{matrix}4&3\\8&7\end{matrix}\right))$$

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

$$3\times 5-7-2\left(4\times 5-8\right)+4\left(4\times 7-8\times 3\right)$$

Simplify.

$$8-2\times 12+4\times 4$$

Add the terms to obtain the final result.

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Factor

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