Solve {ccc}{i}&{j}&{k}{-1}&{-1}&{2}{-3}&{3}&{0}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { - 1 } & { - 1 } & { 2 } \\ { - 3 } & { 3 } & { 0 } \end{array} \right|$$

Evaluate

$$-6i-6k-6j$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\-1&-1&2\\-3&3&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\-1&-1&2&-1&-1\\-3&3&0&-3&3\end{matrix}\right)$$

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

$$j\times 2\left(-3\right)+k\left(-1\right)\times 3=-6j-3k$$

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

$$-3\left(-1\right)k+3\times \left(2i\right)=3k+6i$$

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

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

Subtract 3k+6i from -6j-3k.

$$-6i-6k-6j$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\-1&-1&2\\-3&3&0\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}-1&2\\3&0\end{matrix}\right))-jdet(\left(\begin{matrix}-1&2\\-3&0\end{matrix}\right))+kdet(\left(\begin{matrix}-1&-1\\-3&3\end{matrix}\right))$$

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

$$i\left(-3\times 2\right)-j\left(-\left(-3\times 2\right)\right)+k\left(-3-\left(-3\left(-1\right)\right)\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$-6i-6k-6j$$