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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!

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