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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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