# 3x3 matrix

1. Jan 12, 2016

### Bruno Tolentino

I have a doubt...

Look this matrix equation:
$$\begin{bmatrix} A\\ B \end{bmatrix} = \begin{bmatrix} +\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\ +\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} X\\ Y \end{bmatrix}$$
$$\begin{bmatrix} X\\ Y \end{bmatrix} = \begin{bmatrix} +\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\ +\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} A\\ B \end{bmatrix}$$
By analogy, should exist a matrix 3x3 such that:
$$\begin{bmatrix} A\\ B\\ C\\ \end{bmatrix} = \begin{bmatrix} ? & ? & ?\\ ? & ? & ?\\ ? & ? & ?\\ \end{bmatrix} \begin{bmatrix} X\\ Y\\ Z\\ \end{bmatrix}$$
$$\begin{bmatrix} X\\ Y\\ Z\\ \end{bmatrix} = \begin{bmatrix} ? & ? & ?\\ ? & ? & ?\\ ? & ? & ?\\ \end{bmatrix} \begin{bmatrix} A\\ B\\ C\\ \end{bmatrix}$$
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/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.

2. Jan 12, 2016

### Staff: Mentor

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.
Yes, as long as the matrix has an inverse. The equation I wrote is applicable for any square matrix A that has an inverse.
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.
Of course.

Last edited: Jan 12, 2016
3. Jan 12, 2016

### mathman

The problem in matrix form is find A (if it exist other than A=I) where $A^2=I$. In scalar form it is 9 equations with 9 unknowns.

4. Jan 13, 2016

### Hawkeye18

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.