# Prove A is invertable for all values of theta

#### delgeezee

##### New member
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

##### Active member
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

Staff member
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

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Helper
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 $$A^{-1}(\theta)=A(-\theta)=\begin{bmatrix}\cos(-\theta) & -\sin(-\theta) &0 \\ \sin(-\theta) & \;\;\cos(-\theta) & 0 \\ 0&0&1 \end{bmatrix}=\begin{bmatrix}\;\;\cos\theta & \sin\theta &0 \\ -\sin\theta & \cos\theta & 0 \\ 0&0&1 \end{bmatrix}$$