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Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian

Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.

ISI MS PAPER SOLVED ON LINEAR ALGEBRA

SOLUTION TO ISI SAMPLE PAPER

Here is 5 solved problems from ISI MS PAPER(recently named PSB PAPER) from Linear Algebra.
PROBLEMS ARE TAKEN FROM ISI MS(PSB) SAMPLE PAPER.
Problem.1.Let P be a matrix of order>1 & entries are +ve integers. Suppose inverse of P exists & has integer entries,then what are the set of possible values of |P| ?
Sol:- P has integer values.
=> k1+k2+ . . . . +kn =trace(P)=integer
=> ∑ kikj = Sum of minors of order,here the sum is taken over i<j .
.
.
.
∏ ki = |P| =integer,where i=1(1)n.
Then the eigen values of inverse of P are 1/ki & they are also integers.
=> ki =1/ki
=> ki =+1,-1.
=> |P| = ∏ ki = +1,-1, where i=1(1)n.

Problem.2.Let A and B be two matrices of order p*q & q*p,respectively.Show that |AB|=0 if p>q.
Sol:- AB is a matrix of order p*p.
Rank(AB) ≤ min[ Rank(A),Rank(B) ]
For q<p, Rank(A) ≤ q   &  Rank(B) ≤ q.
Given that Rank(AB) ≤ q.
=> Columns of AB are linearly dependent.
=> |AB|=0.

Problem.3. Let A & B be two invertible n*n matrix. Assume that A+B is invertible. Then Show that A^(-1) + B^(-1) is invertible.
Hints:- |A(A^(-1)+B^(-1))B|=|A+B|
=> A^(-1) + B^(-1) = |A+B|/{|A||B|}.

Problem.4. [JAM-2007] Let  T : R^3 ->R^2 be a linear transformation defined by
T(x,y,z)=(x+y,x-z). Then the dimension of the null space of T is
A.0      B.1      C.2     D.3
Sol:- (B)  So,  (x+y,x-z)=(0,0)
=> x=-y=z.
Let x=a ; N(T)={(a,-a,a)|a€R}.
So, {1,-1,1} is a basis for ker(T).
=> dim(kerT)=1.

Problem.5. Let XY be a bivariate normal vector such that E(X)=E(Y)=0 & V(X)=V(Y)=1. Let S be a subset of R^2 and defined by
S={(a,b) : (ax+by) is independent of Y}.Show that
I. S be a subspace.     II.Find its dimension.
Sol:- I. (a1,b1),(a2,b2) € S.
Then a1x+b1y is independent of Y
And a2x+b2y is independent of Y.
=> (ma1+na2)x + (mb1+nb2)y is indep of Y.
=> (ma1+na2,mb1+nb2) € S
=> m(a1,b1)+n(a2,b2) € S for all (m,n)€R.
Hence, S is a subspace.
II. (a,b) € S
=> ax+by is independent of Y.
=> Cov[ax+by,y]=0
=> aCov[x,y]+bCov[y,y] =0
=> ar + b.1 =0, since Cov[x,y]=R
=> b= -ar, also Cov[y,y]=V[Y]=1
=> (a,b)=a(1,-r) ,a€R.
So, S={(a,b):(a,b)=a(1,-r),a€R}
=> dim(S)=1.