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

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 7+3+2\times 3\times 5=47$$

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

$$2+5\times 3\times 2+7\times 3=53$$

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

$$47-53$$

Subtract 53 from 47.

$$-6$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

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

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.

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

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

$$2\left(7-5\times 3\right)-\left(3\times 7-3\right)+2\left(3\times 5-1\right)$$

Simplify.

$$2\left(-8\right)-18+2\times 14$$

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$