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# Trace & Determinant | ISI MStat 2017 Problem 1 | PSB

This is a beautiful problem from ISI MStat 2017 PSB based on matrices . We provide details solution with the prerequisites mentioned explicilty.

This is a beautiful problem from ISI MStat 2017 PSB Problem 1 based on matrices. We provide a detailed solution with the prerequisites mentioned explicitly.

## Problem – ISI MStat 2017 Problem 1

Let $a$ and $b$ be real numbers. Show that there exists a unique $2 \times 2$ real symmetric matrix $A$ with ${trace}(A)=a$ and $Det(A)=b$ if and only if $a^{2}=4 b$ .

## Prerequisites

• Trace
• Determinant
• Sridhar Acharya Formula

## Solution

Let , $A = \begin{bmatrix} x & z \\ z & y \end{bmatrix}$ be a unique $2 \times 2$ real symmetric matrix , with x,y,z belongs to real .

Given , trace(A)=$a \Rightarrow (x+y)= a$ —-(1)and

Det(A)=$b \Rightarrow (xy-z^2)=b$ —-(2)

1st solution : $x=a-y$ from (1) putting this in (2) we get

$y(a-y)-z^2=b \Rightarrow ay-y^2-z^2=b \Rightarrow y^2-ay+(b+z^2)=0$

Now using , Sridhar Acharya Formula we get ,

$y= \frac{a \pm \sqrt{a^2-4(b+z^2)}}{2}$

Now as given , A is unique so y can’t take two different values hence this part $\sqrt{a^2-4(b+z^2)}$ must be zero

i.e $\sqrt{a^2-4(b+z^2)}=0$ $\Rightarrow z= \pm { \sqrt{\frac{a^2-4b}{4}}}$

Again , as A is unique matrix z can’t take two different values . Hence z must be equal to zero

i.e $z=0 \Rightarrow \sqrt{\frac{a^2-4b}{4}}=0 \Rightarrow a^2=4b$ ( Hence proved)

2nd solution : See if we interchange x by y then all the properties of A remains same i.e trace and determinant . But it can’t be possible as we assume that A is unique hence x and y must be equal i.e $x=y$ .

From (1) we get $x=y=\frac{a}{2}$

Again if we interchange z by -z then all the properties of A remains same i.e trace and determinant . But it can’t be possible as we assume that A is unique hence z and -z must be equal i.e $z=-z \Rightarrow z= \frac{1}{2}$ .

From (2) we get $(xy-z^2)=b \Rightarrow \frac{a^2}{4}-\frac{1}{4} = b \Rightarrow a^2=4b$ (Hence proved )

Now , we will assume that $a^2=4b$ , where a and b are real and show that the matrix is unique .

Let , $A = \begin{bmatrix} x & z \\ z & y \end{bmatrix}$ be a $2 \times 2$ real symmetric matrix with x,y,z belongs to real

Given trace(A)=$a \Rightarrow (x+y)=a$—(3)

and Det(A)=$b \Rightarrow xy-z^2=b$ —(4)

Another thing is given that $a^2=4b$

So using (3) and (4) in (5) we get ,

${(x+y)}^2=4(xy-z^2) \Rightarrow x^2+2xy+y^2-4xy+4z^2=0 \Rightarrow {(x-y)}^2+{(2z)}^2=0$

i.e sum of two squares is equal to 0 which implies individual squares are equal to zero .

Hence , ${(x-y)}^2 =0$ and ${(2z)}^2=0$ $\Rightarrow x=y$ and $z=0$

which give $x=y=\frac{a}{2}$ and $z=0$

Hence , $A= \begin{bmatrix} \frac{a}{2} & 0 \\ 0 &\frac{a}{2} \end{bmatrix}$ is unique ( proved )

we have proved both if and only if part . Hence we are done!

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