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### 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.