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

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

Evaluate

$$23$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 5\times 2+3\times 3+7\times 3\times 7=176$$

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

$$3\times 5\times 7+7\times 3\times 2+2\times 3=153$$

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

$$176-153$$

Subtract 153 from 176.

$$23$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-11\right)-\left(-3\right)+7\times 6$$

Add the terms to obtain the final result.

$$23$$

Factor

$$23$$