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

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!

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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