Solve {lll}{5}&{3}&{3}{6}&{4}&{4}{5}&{1}&{2}

$$\left( \begin{array} { l l l } { 5 } & { 3 } & { 3 } \\ { 6 } & { 4 } & { 4 } \\ { 5 } & { 1 } & { 2 } \end{array} \right)$$

Calculate Determinant

$$2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&3&3\\6&4&4\\5&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&3&3&5&3\\6&4&4&6&4\\5&1&2&5&1\end{matrix}\right)$$

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

$$5\times 4\times 2+3\times 4\times 5+3\times 6=118$$

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

$$5\times 4\times 3+4\times 5+2\times 6\times 3=116$$

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

$$118-116$$

Subtract 116 from 118.

$$2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&3&3\\6&4&4\\5&1&2\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}4&4\\1&2\end{matrix}\right))-3det(\left(\begin{matrix}6&4\\5&2\end{matrix}\right))+3det(\left(\begin{matrix}6&4\\5&1\end{matrix}\right))$$

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

$$5\left(4\times 2-4\right)-3\left(6\times 2-5\times 4\right)+3\left(6-5\times 4\right)$$

Simplify.

$$5\times 4-3\left(-8\right)+3\left(-14\right)$$

Add the terms to obtain the final result.

$$2$$

Evaluate

$$\left(\begin{matrix}5&3&3\\6&4&4\\5&1&2\end{matrix}\right)$$