If $A=\begin {bmatrix} a&b\\b&a \end{bmatrix}$ and $A^2=\begin{bmatrix} x&y \\ y&x \end{bmatrix}$, find min value of $x/y$

Asked May 10 '20 at 06:42

Active May 10 '20 at 06:42

Viewed 45 times

Here,

$$A^2=\begin{bmatrix} a^2+b^2 & 2ab \\ 2ab & a^2+b^2 \end{bmatrix}$$

Now the value of $\frac xy =\frac{a^2+b^2}{2ab}$

From here, if I use $a=b$, I get $\frac xy =1$, which is indeed the right answer.

My problems: If I use $a+b=0$, then I can get $\frac xy =-1$, which is smaller than the given value, and I think more solutions can be found where the answer will be $-2,-4..$ (I haven’t found them, I just have a feeling).

Is the given answer wrong, or is there some special rule that prevents the ratio from being negative?

asked May 10 '20 at 06:42

Aditya

6,191

4

Are you sure that there is no restriction on $a$ and $b$? – José Carlos Santos May 10 '20 at 06:49
@JoséCarlosSantos yes, I am confident – Aditya May 10 '20 at 14:36
Is A supposed to be non singular?bcoz if a=-b then A becomes singular – Atharv May 10 '20 at 15:07
@Atharv no such relation is given – Aditya May 10 '20 at 17:25
Since $(a, b)$ and $(\lambda a, \lambda b)$, $\lambda \neq 0$, give the same ratio $\frac{x}{y}$ for $ab \neq 0$, we may as well take $a = 1$, so that $\frac{x}{y} = \frac{1 + b^2}{2 b}$. But this quantity tends to $-\infty$ as $b$ does, so if there is no restriction on $a, b$ (say, $a, b > 0$, though $a b > 0$ suffices), there is no minimum. – Travis Willse May 10 '20 at 19:04

If $A=\begin {bmatrix} a&b\\b&a \end{bmatrix}$ and $A^2=\begin{bmatrix} x&y \\ y&x \end{bmatrix}$, find min value of $x/y$

0 Answers0