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

  • #1
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.

[tex]
\begin{cases}
x = x' cos \theta - y' sin \theta\\
y = x' sin \theta + y'cos \theta
\end{cases}
[/tex]

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##.

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

[tex]
Let A_1 =
\begin{bmatrix}
x & -sin \theta\\
y & cos\theta
\end{bmatrix}
[/tex]
[tex]
Let A_2 =
\begin{bmatrix}
cos \theta & x\\
sin \theta & y
\end{bmatrix}
[/tex]
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.

[tex]
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}
[/tex]

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

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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.

[tex]
\begin{cases}
x = x' cos \theta - y' sin \theta\\
y = x' sin \theta + y'cos \theta
\end{cases}
[/tex]

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##.

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

[tex]
Let A_1 =
\begin{bmatrix}
x & -sin \theta\\
y & cos\theta
\end{bmatrix}
[/tex]
[tex]
Let A_2 =
\begin{bmatrix}
cos \theta & x\\
sin \theta & y
\end{bmatrix}
[/tex]
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.

[tex]
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}
[/tex]

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.
 
  • #3
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.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
By the way, the original equations,
[itex]x= x'cos(\theta)- y'sin(\theta)[/itex]
[itex]y= x'sin(\theta)+ y'cos(\theta)[/itex]

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

[itex]x'= xcos(-\theta)- y sin(-\theta)= xcos(\theta)+ ysin(\theta)[/itex],
[itex]y'= xsin(-\theta)+ ycos(-\theta)= -x sin(\theta)+ y cos(\theta)[/itex].
 

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