-2
$\begingroup$

I was looking to the exercise for Linear Algebra from my course materials. Stumbled upon this question, and I have no idea how to start or solve this question. As I am totally new to the realm of Linear Algebra, I wanted to solve the question which is mentioned in the exercise of my course materials named "Advanced Mathematics" for the course Master's in Electronics & Telecommunication.

Consider the transformation $T: U_{2\times 2} \rightarrow \mathbb{R}_2[x], \text{ where } U_{2\times 2} = \{ A \in \mathbb{R}_{2 \times 2}, a_{21} = 0 \}$ given by
$T\begin{bmatrix} a & b \\ 0 & c \end{bmatrix}=(x+1)^2(2c-a-b)-(3x-2)(a+b)+x(5c-b)-3c$.

Find the formula for the inverse transformation.

$\endgroup$
7
  • 3
    $\begingroup$ Welcome to MathSE! Unfortunately, your post doesn’t fully meet this site’s policy (see how to ask a good question). I recommend that you edit your post. (1) Please use MathJax to type all your formulae. (2) You should provide some context to your question. Where is this problem from? Why is it interesting and important? (3) Please tell in the post, did you try to solve it? What progress did you achieve? Where were you “stuck”? $\endgroup$
    – Aig
    Apr 12 at 21:52
  • $\begingroup$ Yeah, the style here is to give some context, for sure. It's not meant to be hostile, but just to give some substance to engage-with. :) $\endgroup$ Apr 12 at 22:08
  • $\begingroup$ @paulgarrett I'm also stuck with this. Only this is being given to find the formula for Inverse Transformation. $\endgroup$ Apr 12 at 22:23
  • 4
    $\begingroup$ @Piq, you don't know who is voting to close this question, and you don't know that those voting to close it can't solve it. $\endgroup$ Apr 13 at 1:10
  • 1
    $\begingroup$ @Gerry Myerson: Finally I agree with you. ( I delette my comment)(On more than one occasion, I have tried hard to solve a problem and when I have achieved the solution I have found that the post has been closed). I am sorry. $\endgroup$
    – Piquito
    Apr 14 at 22:33

1 Answer 1

1
$\begingroup$

HINT (before possible closure of the post).- The set of all the matrices $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ is a vectorial space $E$ of dimension $3$ and T apply $E$ in the vectorial space of quadratic trinomial $rx^2+sx+t$ of dimension $3$. The kernel of $T$ is equal to $\{0\}$ so $T$ is bijective (by finite dimension). We have $$T\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}=(2c-a-b)x^2+(9c-5a-6b)x+(a-c)=rx^2+sx+t$$

$$\begin{pmatrix} 2 & -1&-1 \\ 9 & -5&-6\\-1&1&0 \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}r\\s\\t\end{pmatrix}$$ and the determinant of this last matrix is equal to $10$ so for each $r,s,t$ there is a $a,b,c$. The inverse transformation is given by the matrix inverse which is $$\frac12\begin{pmatrix} 6 & -1&1 \\ 6 & -1&3\\4&-1&-1 \end{pmatrix}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .