Solving equation different solutions

Question

The vector $b=\pmatrix{x1\\x2}$ gets roated with a matrix about $30 \deg $

what results in the vector $ \overline b =\pmatrix{6\\8} $

Now my task is to find the original vector coordinates of $b$

I started like this:

The transformation-matrix is $T =\pmatrix{cos(30) & -sin(30) \\ sin(30) & cos(30)}$ and: $$\pmatrix{cos(30) & -sin(30) \\ sin(30) & cos(30)} * \pmatrix{x1\\x2} = \pmatrix{6\\8}$$

What means:

$$x1 *cos(30)+ x2* -sin(30) = 6$$ $$x1*sin(30) + x2 * cos(30)=8$$

Next i dissolved the two equations for $x1 = 1$:

$$x2 = {6-cos(30)\over -sin(30)} $$

$$x2 = {8-sin(30\over cos(30)}$$

When i type this into my calculator i get for the first $x2 = -10.267...$ and for the second equation $x2 = 8.6602...$

My question is: Why do i get different solutions for $x2$? Did i make a mistake or is my calculator the problem? Thanks

Answer 1

You get two different answers because $x_1 = 1$ is not the correct solution. Your two equations define two lines in the plane; they intersect at a single point. When you pick $x_1 = 1$, you get the $x_2$ values for the point that lie on each line with $x$-coordinate 1.

The real problem is that when you "dissolved" the two equations, you did it wrong. The first should say $$ x_2 = \frac{6 - x_1 \cos 30}{-\sin 30} $$ Now you should write out the second one in a similar form. That gives two different expressions for $x_2$, which you can set equal to each other, and solve for $x_1$, and you'll be on your way.

Following your comment:

You cannot "do it for the first equation": you need to solve the two equations together. So far we have $$ x_2 = \frac{6 - x_1 \cos 30}{-\sin 30} $$ For the second -- $$ x_1 \sin 30 + x_2 \cos 30 = 8 $$ -- you can do the following steps: \begin{align} x_1 \sin 30 + x_2 \cos 30 &= 8\\ x_2 \cos 30 &= 8 - x_1 \sin 30 \\ x_2 &= \frac{8 - x_1 \sin 30}{\cos 30}. \end{align} Now we have two expressions for $x_2$ which we can set equal: \begin{align} x_2 &= \frac{8 - x_1 \sin 30}{\cos 30} \text{, and}\\ x_2 &= \frac{6 - x_1 \cos 30}{-\sin 30}\text{, so}\\ \frac{8 - x_1 \sin 30}{\cos 30} &= \frac{6 - x_1 \cos 30}{-\sin 30}.\\ (8 - x_1 \sin 30)(-\sin 30)&= (6 - x_1 \cos 30)({\cos 30} )\\ -8\sin 30 + x_1 \sin^2 30&= 6\cos 30 - x_1 \cos^2 30\\ x_1 \sin^2 30 + x_1 \cos^2 30&= 6\cos 30 + 8 \sin 30\\ x_1 (\sin^2 30 + \cos^2 30)&= 6\cos 30 + 8 \sin 30\\ x_1 &= 6\cos 30 + 8 \sin 30\\ ~\approx 9.2. \end{align}

Similarly, you can find that $$ x_2 = -6 \sin 30 + 8 \cos 30 \approx 3.93 $$

Note that these final computations of $x_1$ and $x_2$ are exactly those suggested in @user84413's answer; I worked out these details because I thought you might not be that familiar with the matrix mathematics yet. That other answer is clearly the way to go, once you get more familiar with stuff like this.

Answer 2

Another way you could approach this problem, which might be a little easier, is to use the fact that the inverse matrix will rotate $\pmatrix{6\\8}$ back to $\pmatrix{x_{1}\\x_{2}}$.

Since the inverse matrix corresponds to a rotation of $-30^{\circ}$, you have that

$\pmatrix{x_{1}\\x_{2}}=\begin{pmatrix} \cos(-30^{\circ}) & -\sin(-30^{\circ})\\ \sin(-30^{\circ}) & \cos(-30^{\circ})\end{pmatrix}\pmatrix{6\\8}$

where $\cos(-30^{\circ})=\cos(30^{\circ})$ and $\sin(-30^{\circ})=-\sin(30^{\circ})$.