Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the matrix representation $A$ of a linear mapping $T$ : $\mathbb R^2 \to \mathbb R^2$ that rotates points $\pi$ radians clockwise, then reflects points through the line $x_2 = - x_1$. Determine the range of $T$. Determine if $(1,1)$ is in the range of $T$. Determine if this linear mapping $T$ is invertible by calculating the determinant of $A$.

Alright, I've been trying to solve the following problem for nearly 3 hours now, but no avail. I don't even fully understand the question. I didn't find any tutorials online that met my needs. I'd appreciate if someone could go through the solution with (brief if possible) explanation of the problem.

share|improve this question
Not sure what's confusing you. Do you know the matrix representation of a rotation? Of a reflection? If you know both of those, then all that remains is just to calculate the overall matrix of the transformation through multiplication. –  Muphrid Jun 27 '13 at 5:19
This problem belongs to a B-term course. I have a really bad teacher who'd expect everyone in class to study and learn on their own. i would really appreciate if you show me breifly how it's done, or redirect me to an online tutorial that does solve a very similar problem. –  Ryan Hanna AL-Kass Jun 27 '13 at 5:24
I have no basis for where to begin. I could start talking about matrices that represent rotations, but you may not find that obvious. Please explain in more detail what about this problem is confusing to you, so I can target an answer to those issues. –  Muphrid Jun 27 '13 at 5:27
okay so if the inverse of the matrix is not equal to 0 (zero), the matrix is invertible? right? how do I come up with the matrix itself? also, how do i determine of (1, 1) is (or is NOT) in the range of T? the books has a lot of explanation that none of it seems to make any sense to me –  Ryan Hanna AL-Kass Jun 27 '13 at 5:32
Yeah, there are a lot of little steps that we really need to know your understanding of before trying to give a full answer. I suppose I'd start with: 1. Can you write the matrices that represent the two individual transformations (rotation and reflection)? 2. Given 1, can you explain how you can use those matrices to apply the transformations to a generic vector? 3. Based on that, can you explain what you would do with the two matrices to apply the combined transformation $T$ as described? 4. Combining those, can you work out what $A$ would be? –  ConMan Jun 27 '13 at 5:34
add comment

2 Answers

up vote 1 down vote accepted

It is sufficient to see how $T$ behaves on a basis of $\mathbb{R}^2$.

Without doing any computation, you can guess that $T$ is invertible, since both operations of reflecting through the line, and rotating by $\pi$ are invertible.

$T$ takes the point $e_1=(1,0)^T$, rotates by $\pi$ to get $(-1,0)^T$ and reflects in the line $L$ to get $e_2=(0,1)^T$. ($L$ is the line $y=-x$.)

$T$ takes the point $e_2=(0,1)^T$, rotates by $\pi$ to get $(0,-1)^T$ and reflects in the line $L$ to get $e_1=(1,0)^T$.

Hence the representation of $T$ in terms of the basis $e_1,e_2$ is given by: $A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.

To see if $(1,1)^T$ is in the range, solve $A(x,y)^T = (1,1)^T$, which quickly gives $x=y=1$, so yes, the point is in the range.

In fact, it is clear from this that the equation $A(x,y)^T = (a,b)^T$ has the solution $(x,y)^T = (b,a)^T$, for any $a,b$, hence $T$ (equivalently, $A$) is invertible.

The determinant is straightforward to compute as $\det A = 0 \cdot 0 - 1 \cdot 1 = -1$.

share|improve this answer
add comment

$A$ is composed of a rotation $R$ and of a reflection $M$, $A=MR$. The rotation is the rotation matrix with the angle set to $\pi$, but if you don't know what this looks like, you can guess: $R$ must take $(1,0)$ to $(-1,0)$, and $(1,1)$ to $(-1,-1)$, i.e: $$R = \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array} \right)$$ As to the reflection, this takes nay point $(x,y)$ to $(-x,y)$, or in other "words": $$M = \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)$$ Multiply to get $A$, and if the determinant is not zero, then $A$ is invertible.

share|improve this answer
I think you need $M=\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$? –  copper.hat Jun 27 '13 at 5:49
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.