# How find this matrix in an equation

Let $x$ is give numbers.and define the matrix $A$ such $$A^T\cdot\begin{bmatrix} 1&x\\ x&1 \end{bmatrix}\cdot A=1$$

this book say it is clear have $$A=\dfrac{1}{\sqrt{2(1+x)}}\begin{bmatrix} 1\\ 1 \end{bmatrix}$$

I think this not true,let $$A=\begin{bmatrix} a\\ b \end{bmatrix}$$ then we have $$A^T\cdot\begin{bmatrix} 1&x\\ x&1 \end{bmatrix}\cdot A=\binom{a+bx}{b+ax}\cdot\begin{bmatrix} a\\ b \end{bmatrix}=a^2+b^2+2abx=1$$

First which is wrong?

why is "clear" can you explain? Thank you,If not it is clear,can you solve this equation

• Is there some additional condition that means that $A = (1\ \ 0)^T$ doesn't work? – Ben Millwood Jan 4 '15 at 12:56
• It seems like the given answer does work and is the only answer such that $a = b$. Is that necessary for some reason? Or is it just the case that the book is trying to give a solution rather than the solution? – Ben Millwood Jan 4 '15 at 12:59
• oh,maybe this book is not true, if $a=b$ is true – china math Jan 4 '15 at 13:07

if you let $$A=\begin{bmatrix} a\\ b \end{bmatrix}$$ then we have $$A^T\cdot\begin{bmatrix} 1&x\\ x&1 \end{bmatrix}\cdot A=\binom{a+bx}{b+ax}\cdot\begin{bmatrix} a\\ b \end{bmatrix}=a^2+b^2+2abx=1$$
the book assumed a case where $a=b$.
This will then give $a^2+a^2+2a^2x=1$ and solving for a gives: $$a=\dfrac{1}{\sqrt{2(1+x)}}$$ Therefore we will have $$A=\dfrac{1}{\sqrt{2(1+x)}}\begin{bmatrix} 1\\ 1 \end{bmatrix}$$