#### 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 .

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.

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∏ 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.