# Linear Algebra, Cramers Rule - Solving for unknowns in a matrix

1. Mar 11, 2013

### johnstobbart

Hello everyone, I have a linear algebra question regarding Cramer's rule.

1. The problem statement, all variables and given/known data

Using Cramer's rule, solve for x' and y' in terms of x and y.

$$\begin{cases} x = x' cos \theta - y' sin \theta\\ y = x' sin \theta + y'cos \theta \end{cases}$$

2. Relevant equations

$sin^2 \theta + cos^2 \theta = 1$

3. The attempt at a solution

I need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing $x_1$ and $x_2$.

$$Let A = \begin{bmatrix} cos \theta & -sin \theta\\ sin \theta & cos\theta \end{bmatrix}$$
I form two more matrices, $A_1$ and $A_2$.

$$Let A_1 = \begin{bmatrix} x & -sin \theta\\ y & cos\theta \end{bmatrix}$$
$$Let A_2 = \begin{bmatrix} cos \theta & x\\ sin \theta & y \end{bmatrix}$$
I then find $det(A)$. I get $cos^2 \theta + sin^2 \theta$ which is $1$.
$det(A_1) = x cos \theta + y sin \theta$
$det(A_2) = y cos \theta - x sin \theta$
Lastly, I need to find the value of $x'$ and $y'$, using Cramer's Rule.

$$x' = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\ y' = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}$$

Can anyone tell me if I'm on the right track for this problem?

Last edited: Mar 11, 2013
2. Mar 11, 2013

### Ray Vickson

A good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.

3. Mar 12, 2013

### johnstobbart

Ray, I completely forgot about being able to check questions like these. I want to sincerely thank you for reminding me of this rule.

4. Mar 12, 2013

### HallsofIvy

Staff Emeritus
By the way, the original equations,
$x= x'cos(\theta)- y'sin(\theta)$
$y= x'sin(\theta)+ y'cos(\theta)$

give the (x, y) coordinates of point (x', y') after a rotation through angle $\theta$, clockwise, around the origin. Going from (x, y) back to (x', y') is just a rotation in the opposite direction. So

$x'= xcos(-\theta)- y sin(-\theta)= xcos(\theta)+ ysin(\theta)$,
$y'= xsin(-\theta)+ ycos(-\theta)= -x sin(\theta)+ y cos(\theta)$.