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

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

Evaluate

$$6j+8k-4i$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$j\times 2\times 3+k\times 2=6j+2k$$

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

$$3\left(-2\right)k+2\times \left(2i\right)=4i-6k$$

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

$$6j+2k-\left(4i-6k\right)$$

Subtract -6k+4i from 6j+2k.

$$6j+8k-4i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-4i-j\left(-6\right)+k\times 8$$

Add the terms to obtain the final result.

$$6j+8k-4i$$