Solve {lll}{i}&{i}&{k}{3}&{1}&{2}{2}&{-2}&{4}

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

Evaluate

$$-8k$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4i+2i\times 2+k\times 3\left(-2\right)=8i-6k$$

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

$$2k-2\times \left(2i\right)+4\times \left(3i\right)=2k+8i$$

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

$$8i-6k-\left(2k+8i\right)$$

Subtract 2k+8i from 8i-6k.

$$-8k$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$8i-8i+k\left(-8\right)$$

Add the terms to obtain the final result.

$$-8k$$