2

I have a problem to solve, and it makes me realise that I have a bad understanding of basis change and the linear applications that go along. I have a linear application who's matrix is expressed like this:

$A=\begin{bmatrix} \frac{4a-b}{3}&\frac{2a-2b}{3} \\ \frac{-2a+2b}{3}& \frac{-a+4b}{3} \end{bmatrix} $

I suppose in order to get the basis of this linear appplication I should compute the values of $Av_{1}$ and $Av_{2}$ where $v_{1}=\binom{1}{0}$ and $v_{2}=\binom{0}{1}$. Am I correct so far?

Now, where my lack of understand lacks is here: how to I determine the matrix $A^{'}$ that is the associated matrix in the base $x_{1} \:x_{2}$? Should I just compute $Ax_{1}$ and $Ax_{2}$?

EDIT 1: I made a small mistake in the matrix: it's not$$A=\begin{bmatrix} \frac{4a-b}{3}&\frac{2a-2b}{3} \\ \frac{-2a+2b}{3}& \frac{-a+4b}{3} \end{bmatrix} $$ but $$A=\begin{bmatrix} \frac{4a-b}{3}&\frac{-2a+2b}{3} \\ \frac{2a-2b}{3}& \frac{-a+4b}{3} \end{bmatrix} $$

I calculated the determinant of this matrix, and found that it is equal to: $3ab$. I know that to find the inverse of the matrix A, I can use the adjudant matrix. So normally I should be able to compute this: $$A^{-1}=\frac{1}{3ab}*adj(A)$$

For $adj(A)$ I did this:

  • Find the cofactor matrix:

$Com(A)$: $$Com(A)=\begin{bmatrix} \frac{-a+4b}{3}&\frac{2a-2b}{3} \\ \frac{2b-2a}{3}& \frac{4a-b}{3} \end{bmatrix} $$

The adjugate matrix will be equal to the transpose of the $Com(A)$. So we have: $$adj(A)=\begin{bmatrix} \frac{-a+4b}{3}&\frac{2b-2a}{3} \\ \frac{2a-2b}{3}& \frac{4a-b}{3} \end{bmatrix} $$

  • Find the expression of the inverse:

I have now: $$A^{-1}=\frac{1}{3ab}*\begin{bmatrix} \frac{-a+4b}{3}&\frac{2b-2a}{3} \\ \frac{2a-2b}{3}& \frac{4a-b}{3} \end{bmatrix}$$

Or in a simplified way:

$$A^{-1}=\frac{1}{9ab}*\begin{bmatrix} -a+4b&2b-2a\\ 2a-2b& 4a-b \end{bmatrix}$$

So now all I have to do is to chose the two vectors I want $v_{1},v_{2}$ and calculate $A^{-1}v_{1}$ and $A^{-1}v_{2}$? And this will give me the basis of the linear transformation in the basis ($v_{1},v_{2}$), right?

EDIT 2:

I made a mistake. I should not calculate $A^{-1}v_{1}$ and $A^{-1}v_{2}$ but solve $v_{1}=A^{-1}\begin{bmatrix} \frac{4a-b}{3}\\ \frac{2a-2b}{3} \end{bmatrix}$ and $v_{2}=A^{-1}\begin{bmatrix} \frac{-2a-2b}{3}\\ \frac{-a+4b}{3} \end{bmatrix}$

John Mayne
  • 2,148

1 Answers1

1

HINT.-The determinant of $A$ is equal to $ab$ so if $ab\ne 0$ then the matrix is invertible. If you want to know the base $\{b_1,b_2\}$ where $b_1=\begin{pmatrix} x_1\\y_1\end{pmatrix}$ and $b_2=\begin{pmatrix} x_2\\y_2\end{pmatrix}$ you have to solve

$$A\begin{pmatrix} x_1\\y_1\end{pmatrix}=\begin{pmatrix} \frac{4a-b}{3}\\\frac{-2a+2b}{3}\end{pmatrix}$$

$$A\begin{pmatrix} x_2\\y_2\end{pmatrix}=\begin{pmatrix} \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$ For this you multiply both equations for the inverse $A^{-1}$ of $A$ (you know that if $ab\ne 0$ then $A$ is invertible). Finally you have $$\begin{pmatrix} x_1\\y_1\end{pmatrix}=A^{-1}\begin{pmatrix} \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$ $$\begin{pmatrix} x_2\\y_2\end{pmatrix}=A^{-1}\begin{pmatrix} \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$

Do you know calculate the inverse matrix of A?

Piquito
  • 29,594