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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.

Let and be real numbers. Show that there exists a unique real symmetric matrix with and if and only if .

- Trace
- Determinant
- Sridhar Acharya Formula

Let , be a unique real symmetric matrix , with x,y,z belongs to real .

Given , trace(A)= ----(1)and

Det(A)= ----(2)

**1st solution** : from (1) putting this in (2) we get

Now using , Sridhar Acharya Formula we get ,

Now as given , A is unique so y can't take two different values hence this part must be zero

i.e

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

i.e ( 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 .

From (1) we get

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 .

From (2) we get (Hence proved )

Now , we will assume that , where a and b are real and show that the matrix is unique .

Let , be a real symmetric matrix with x,y,z belongs to real

Given trace(A)=---(3)

and Det(A)= ---(4)

Another thing is given that

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

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

Hence , and and

which give and

Hence , 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.

Let and be real numbers. Show that there exists a unique real symmetric matrix with and if and only if .

- Trace
- Determinant
- Sridhar Acharya Formula

Let , be a unique real symmetric matrix , with x,y,z belongs to real .

Given , trace(A)= ----(1)and

Det(A)= ----(2)

**1st solution** : from (1) putting this in (2) we get

Now using , Sridhar Acharya Formula we get ,

Now as given , A is unique so y can't take two different values hence this part must be zero

i.e

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

i.e ( 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 .

From (1) we get

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 .

From (2) we get (Hence proved )

Now , we will assume that , where a and b are real and show that the matrix is unique .

Let , be a real symmetric matrix with x,y,z belongs to real

Given trace(A)=---(3)

and Det(A)= ---(4)

Another thing is given that

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

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

Hence , and and

which give and

Hence , is unique ( proved )

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

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