Solve {ccc}{i}&{j}&{k}{-3}&{-3}&{3}{0}&{-2}&{3}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { - 3 } & { - 3 } & { 3 } \\ { 0 } & { - 2 } & { 3 } \end{array} \right|$$

Evaluate

$$9j+6k-3i$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\-3&-3&3\\0&-2&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\-3&-3&3&-3&-3\\0&-2&3&0&-2\end{matrix}\right)$$

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

$$-3i\times 3+k\left(-3\right)\left(-2\right)=6k-9i$$

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

$$-2\times \left(3i\right)+3\left(-3\right)j=-6i-9j$$

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

$$6k-9i-\left(-6i-9j\right)$$

Subtract -6i-9j from -9i+6k.

$$9j+6k-3i$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\-3&-3&3\\0&-2&3\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}-3&3\\-2&3\end{matrix}\right))-jdet(\left(\begin{matrix}-3&3\\0&3\end{matrix}\right))+kdet(\left(\begin{matrix}-3&-3\\0&-2\end{matrix}\right))$$

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

$$i\left(-3\times 3-\left(-2\times 3\right)\right)-j\left(-3\right)\times 3+k\left(-3\right)\left(-2\right)$$

Simplify.

$$-3i-j\left(-9\right)+k\times 6$$

Add the terms to obtain the final result.

$$9j+6k-3i$$