Solve {lll}{3}&{-2}&{3}{3}&{5}&{-3}{1}&{4}&{4}=

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

Evaluate

$$147$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 5\times 4-2\left(-3\right)+3\times 3\times 4=102$$

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

$$5\times 3+4\left(-3\right)\times 3+4\times 3\left(-2\right)=-45$$

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

$$102-\left(-45\right)$$

Subtract -45 from 102.

$$147$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 32-\left(-2\times 15\right)+3\times 7$$

Add the terms to obtain the final result.

$$147$$

Factor

$$3\times 7^{2}$$