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

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

## Homework Statement

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. Homework Equations

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

## 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:

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Ray Vickson
Homework Helper
Dearly Missed
Hello everyone, I have a linear algebra question regarding Cramer's rule.

## Homework Statement

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. Homework Equations

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

## 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?
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.

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

HallsofIvy
Homework Helper
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

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