Solve {lll}{i}&{j}&{k}{1}&{0}&{2}{3}&{1}&{2}

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

Evaluate

$$4j+k-2i$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2i+2j=2j+2i$$

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

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

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

$$4j+k-2i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$4j+k-2i$$