Prove A is invertable for all values of theta

[delgeezee]

Show that matrix A is invertible for all values of \(\displaystyle \theta\); then find \(\displaystyle A^{-1}\) using
\(\displaystyle A^{-1}= \frac{1}{det(A)}adj(A)\)

A =

cos\(\displaystyle \theta\)	-sin\(\displaystyle \theta\)	0
sin\(\displaystyle \theta\)	cos\(\displaystyle \theta\)	0
0	0	1

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By cofactoring along the 3rd row, I find det(A) = (1)*(\(\displaystyle cos^2\theta + sin^2\theta\)) =1 , which is a nonzero and implies that A is invertible.
To get the Andjuct of A or Adj(A) , I form a cofactor matrix C and transpose it.

Adj(A) = \(\displaystyle C^{T}\) =

A =

cos\(\displaystyle \theta\)	sin\(\displaystyle \theta\)	0
-sin\(\displaystyle \theta\)	cos\(\displaystyle \theta\)	0
0	0	1

which also happens to be \(\displaystyle A^{-1}\)
My Question is how am I suppose to prove A is invertible for all Values of \(\displaystyle /theta\)?

My gut tells me I am suppose to state that \(\displaystyle A^{-1}= \frac{1}{det(A)}adj(A)\) does not depend on \(\displaystyle \theta\). Is there a more definitive way of showing A is invertible for all Values of \(\displaystyle \theta\)?

 

[M R]

My Question is how am I suppose to prove A is invertible for all Values of \(\displaystyle /theta\)?

You found that $$\det(A) \ne 0$$. That's all you need to do.

 

[Jameson]

I agree that you've done enough, but to show that the determinant is non-zero for all $\theta$, I would say something along the lines of the following.

$\forall \theta \in \mathbb{R}$ it follows that $\sin (\theta ) \text{ and} \cos( \theta ) \in \mathbb{R}$. Every element in $A$ is in $\mathbb{R}$ so operations involving those elements are also in $\mathbb{R}$ and the result you found holds for all $\theta$.

Not the best wording maybe, but that's the general idea.

 

[Opalg]

My Question is how am I suppose to prove A is invertible for all Values of \(\displaystyle /theta\)?

My gut tells me I am suppose to state that \(\displaystyle A^{-1}= \frac{1}{det(A)}adj(A)\) does not depend on \(\displaystyle \theta\). Is there a more definitive way of showing A is invertible for all Values of \(\displaystyle \theta\)?

$A^{-1}= \begin{bmatrix}\cos\theta & \sin\theta &0 \\ -\sin\theta & \cos\theta & 0 \\ 0&0&1 \end{bmatrix}$, which does depend on $\theta$. But $\det A=1$, and the constant $1$ is nonzero and does not depend on $\theta$. That is all you need, to conclude that $A$ is invertible for all $\theta.$

 

[Fernando Revilla]

Show that matrix A is invertible for all values of \(\displaystyle \theta\); then find \(\displaystyle A^{-1}\) using
\(\displaystyle A^{-1}= \frac{1}{det(A)}adj(A)\)

Another way of finding $A^{-1}$ in this case (I don't know if you have covered it) is to consider $$A(\theta)= \begin{bmatrix}\cos\theta & -\sin\theta &0 \\ \sin\theta & \;\;\cos\theta & 0 \\ 0&0&1 \end{bmatrix}$$ as a rotation around the $z$-axis by angle $\theta$. Then, by geometric considerations $A(\theta)A(-\theta)=A(0)=I$, so $$A^{-1}(\theta)=A(-\theta)=\begin{bmatrix}\cos(-\theta) & -\sin(-\theta) &0 \\ \sin(-\theta) & \;\;\cos(-\theta) & 0 \\ 0&0&1 \end{bmatrix}=\begin{bmatrix}\;\;\cos\theta & \sin\theta &0 \\ -\sin\theta & \cos\theta & 0 \\ 0&0&1 \end{bmatrix}$$