Solve {ccc}{i}&{j}&{k}{6}&{-3}&{2}{4}&{-1}&{2}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { 6 } & { - 3 } & { 2 } \\ { 4 } & { - 1 } & { 2 } \end{array} \right|$$

Evaluate

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

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\6&-3&2\\4&-1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\6&-3&2&6&-3\\4&-1&2&4&-1\end{matrix}\right)$$

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

$$-3i\times 2+j\times 2\times 4+k\times 6\left(-1\right)=8j-6k-6i$$

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

$$4\left(-3\right)k-2i+2\times 6j=12j-12k-2i$$

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

$$8j-6k-6i-\left(12j-12k-2i\right)$$

Subtract -12k-2i+12j from -6i+8j-6k.

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

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\6&-3&2\\4&-1&2\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}-3&2\\-1&2\end{matrix}\right))-jdet(\left(\begin{matrix}6&2\\4&2\end{matrix}\right))+kdet(\left(\begin{matrix}6&-3\\4&-1\end{matrix}\right))$$

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

$$i\left(-3\times 2-\left(-2\right)\right)-j\left(6\times 2-4\times 2\right)+k\left(6\left(-1\right)-4\left(-3\right)\right)$$

Simplify.

$$-4i-j\times 4+k\times 6$$

Add the terms to obtain the final result.

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