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

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

Evaluate

$$-47$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 7+2\times 2\times 5+3\times 3=57$$

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

$$5\times 4\times 3+2+7\times 3\times 2=104$$

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

$$57-104$$

Subtract 104 from 57.

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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).

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

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.

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

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

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

Simplify.

$$26-2\times 11+3\left(-17\right)$$

Add the terms to obtain the final result.

$$-47$$

Factor

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