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

[johnstobbart]

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.

<br /> \begin{cases}<br /> x = x&#039; cos \theta - y&#039; sin \theta\\<br /> y = x&#039; sin \theta + y&#039;cos \theta<br /> \end{cases}<br />

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 $x_2$.

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

<br /> Let A_1 = <br /> \begin{bmatrix}<br /> x &amp; -sin \theta\\<br /> y &amp; cos\theta<br /> \end{bmatrix}<br />
<br /> Let A_2 = <br /> \begin{bmatrix}<br /> cos \theta &amp; x\\<br /> sin \theta &amp; y<br /> \end{bmatrix}<br />
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.

<br /> x&#039; = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\<br /> y&#039; = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}<br />

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

 

[Ray Vickson]

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.

<br /> \begin{cases}<br /> x = x&#039; cos \theta - y&#039; sin \theta\\<br /> y = x&#039; sin \theta + y&#039;cos \theta<br /> \end{cases}<br />

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 $x_2$.

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

<br /> Let A_1 = <br /> \begin{bmatrix}<br /> x &amp; -sin \theta\\<br /> y &amp; cos\theta<br /> \end{bmatrix}<br />
<br /> Let A_2 = <br /> \begin{bmatrix}<br /> cos \theta &amp; x\\<br /> sin \theta &amp; y<br /> \end{bmatrix}<br />
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.

<br /> x&#039; = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\<br /> y&#039; = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}<br />

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.

 

[johnstobbart]

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]

By the way, the original equations,
x= x&#039;cos(\theta)- y&#039;sin(\theta)
y= x&#039;sin(\theta)+ y&#039;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&#039;= xcos(-\theta)- y sin(-\theta)= xcos(\theta)+ ysin(\theta),
y&#039;= xsin(-\theta)+ ycos(-\theta)= -x sin(\theta)+ y cos(\theta).