3x3 matrix

  • #1
I have a doubt...

Look this matrix equation:
[tex]\begin{bmatrix}
A\\
B
\end{bmatrix} = \begin{bmatrix}
+\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\
+\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix} \begin{bmatrix}
X\\
Y
\end{bmatrix}[/tex]
[tex]\begin{bmatrix}
X\\
Y
\end{bmatrix} = \begin{bmatrix}
+\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\
+\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix} \begin{bmatrix}
A\\
B
\end{bmatrix}[/tex]
By analogy, should exist a matrix 3x3 such that:
[tex]\begin{bmatrix}
A\\
B\\
C\\
\end{bmatrix} = \begin{bmatrix}
? & ? & ?\\
? & ? & ?\\
? & ? & ?\\
\end{bmatrix} \begin{bmatrix}
X\\
Y\\
Z\\
\end{bmatrix}[/tex]
[tex]\begin{bmatrix}
X\\
Y\\
Z\\
\end{bmatrix} = \begin{bmatrix}
? & ? & ?\\
? & ? & ?\\
? & ? & ?\\
\end{bmatrix} \begin{bmatrix}
A\\
B\\
C\\
\end{bmatrix}[/tex]
So, what values need be replaced in ? for the matrix equation above be right?

I think that my doubt is related with these wikipages:
https://en.wikipedia.org/wiki/Quadratic_formula#By_Lagrange_resolvents
https://en.wikipedia.org/wiki/Cubic_function#Lagrange.27s_method
https://en.wikipedia.org/wiki/Quartic_function#Solving_by_Lagrange_resolvent

EDIT: The inverse of the 3x3 matrix need be equal to itself.
 

Answers and Replies

  • #2
34,006
5,660
I have a doubt...

Look this matrix equation:
[tex]\begin{bmatrix}
A\\
B
\end{bmatrix} = \begin{bmatrix}
+\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\
+\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix} \begin{bmatrix}
X\\
Y
\end{bmatrix}[/tex]
[tex]\begin{bmatrix}
X\\
Y
\end{bmatrix} = \begin{bmatrix}
+\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\
+\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{bmatrix} \begin{bmatrix}
A\\
B
\end{bmatrix}[/tex]
Your question is not very complicated. If ##\vec{x} = A\vec{b}##, and A is an invertible matrix, then ##\vec{b} = A^{-1}\vec{x}##. Note that capital letters are usually used for matrix names, and lower case letters are used for vectors or the components of vectors.
Bruno Tolentino said:
By analogy, should exist a matrix 3x3 such that:
[tex]\begin{bmatrix}
A\\
B\\
C\\
\end{bmatrix} = \begin{bmatrix}
? & ? & ?\\
? & ? & ?\\
? & ? & ?\\
\end{bmatrix} \begin{bmatrix}
X\\
Y\\
Z\\
\end{bmatrix}[/tex]
[tex]\begin{bmatrix}
X\\
Y\\
Z\\
\end{bmatrix} = \begin{bmatrix}
? & ? & ?\\
? & ? & ?\\
? & ? & ?\\
\end{bmatrix} \begin{bmatrix}
A\\
B\\
C\\
\end{bmatrix}[/tex]
Yes, as long as the matrix has an inverse. The equation I wrote is applicable for any square matrix A that has an inverse.
Bruno Tolentino said:
None of these links is helpful, as far as I can see. Any linear algebra textbook will have a section on finding the inverse of a square matrix.
Bruno Tolentino said:
EDIT: The inverse of the 3x3 matrix need be equal to itself.
Of course.
 
Last edited:
  • #3
mathman
Science Advisor
7,858
446
The problem in matrix form is find A (if it exist other than A=I) where [itex]A^2=I[/itex]. In scalar form it is 9 equations with 9 unknowns.
 
  • #4
177
61
You need to find matrices such that ##A^2=I##, or equivalently, such that ##A^{-1}=A##. Any such matrix can be represented as ##A=SJS^{-1}##, where ##S## is an invertible matrix, and ##J## is a diagonal matrix with entries ##\pm 1## on the diagonal . When all diagonal entries of ##J## are ##1## you get ##A=I##, when all equal ##-1## you get ##A=-I##; when ##J## has both ##1## and ##-1## on the diagonal you will get a non-trivial example of a matrix ##A##.

This is a complete description, meaning that any matrix ##A## such that ##A^{-1}=A## can be represented as ##A=SJS^{-1}## (but the representation is not unique). And this is true in all dimensions, not just in dimension 3.
 
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