Solve {rrr}{1}&{-3}&{3}{2}&{1}&{1}{3}&{-2}&{7}

$$\left| \begin{array} { r r r } { 1 } & { - 3 } & { 3 } \\ { 2 } & { 1 } & { 1 } \\ { 3 } & { - 2 } & { 7 } \end{array} \right|$$

Evaluate

$$21$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$7-3\times 3+3\times 2\left(-2\right)=-14$$

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

$$3\times 3-2+7\times 2\left(-3\right)=-35$$

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

$$-14-\left(-35\right)$$

Subtract -35 from -14.

$$21$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$9-\left(-3\times 11\right)+3\left(-7\right)$$

Add the terms to obtain the final result.

$$21$$

Factor

$$3\times 7$$