Question

Write the following transformation in matrix form
$x_1=\frac{\sqrt{3}}{2}{y}_1+\frac{1}{2}{y}_2;x_2=-\frac{1}{2}{y}_1+\frac{\sqrt{3}}{2}{y}_2$.
Hence find the transformation in matrix form which expresses $y_1,y_2$ in terms of $x_1,x_2$.

• A. $y_1=\frac{\sqrt{3}}{2}{x}_1+\frac{1}{2}{x}_2;y_2=\frac{1}{2}{x}_1+\frac{\sqrt{3}}{2}{x}_2$
• B. $y_1=\frac{\sqrt{3}}{2}{x}_1-\frac{1}{2}{x}_2;y_2=\frac{1}{2}{x}_1+\frac{\sqrt{3}}{2}{x}_2$
• C. $y_1=\frac{\sqrt{3}}{2}{x}_1-\frac{1}{2}{x}_2;y_2=\frac{1}{2}{x}_1-\frac{\sqrt{3}}{2}{x}_2$
• D. None of these

Answer 1

The correct option is B $y_1=\frac{\sqrt{3}}{2}{x}_1-\frac{1}{2}{x}_2;y_2=\frac{1}{2}{x}_1+\frac{\sqrt{3}}{2}{x}_2$

$x_1=\frac{\sqrt{3}}{2}{y}_1+\frac{1}{2}{y}_2$ and $x_2=\frac{-1}{2}{y}_1+\frac{\sqrt{3}}{2}{y}_2$
We observe $\frac{\sqrt{3}}{2}{x}_1-\frac{1}{2}{x}_2=\frac{3}{4}{y}_1+\frac{\sqrt{3}}{2}.\frac{1}{2}{y}_2+\frac{1}{4}{y}_1-\frac{\sqrt{3}}{2}\frac{1}{2}{y}_2$
$\Rightarrow \frac{\sqrt{3}}{2}{x}_1-\frac{1}{2}{x}_2=y_1$
Similarly $\frac{1}{2}{x}_1+\frac{\sqrt{3}}{2}{x}_2=\frac{1}{4}{y}_2+\frac{3}{4}{y}_2=y_2$
$\therefore y_1=\frac{\sqrt{3}}{2}{x}_1-\frac{1}{2}{x}_2;y_2=\frac{1}{2}{x}_1+\frac{\sqrt{3}}{2}{x}_2$