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

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

Evaluate

$$3j-4k+6i$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

$$3j-4k-\left(-6i\right)$$

Subtract -6i from 3j-4k.

$$3j-4k+6i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$6i-j\left(-3\right)+k\left(-4\right)$$

Add the terms to obtain the final result.

$$3j-4k+6i$$