CSIR -NET Linear algebra Solved question : part B : June 2016

1. Given a $n\times n$ matrix $B$ defined by $e^B$ by 
$$e^B=\sum_{j=0}^\infty \frac{B^j}{j!}$$
Let p be the characteristic polynomial of $B$ Then the matrix $e^{P(B)}$ is

1. $I_{n\times n}$
2. $0_{n\times n}$
3. $e \times I_{n\times n}$
4. $\pi \times I_{n\times n}$

Solution:

Since every characteristic polynomial is satisfied by the matrix. (caley hamilton Theorem)

So, \(P(B)=0\) So , \(e^0=I_{n\times n}\)

So, 1st option is correct.

2. Let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)\in \mathbb{R}^3 $ be linearly independent. 
Let 
$$\delta_1 =x_2y_3-y_2x_3$$ 
$$\delta_2 =x_1y_3-y_1x_3$$ 
$$\delta_3 =x_1y_2-y_1x_2$$ 
 If V is the span of $x,y$ then,

1. $V=\{(u,v,w):\delta_1u-\delta_2v+\delta_3w=0\}$
2. $V=\{(u,v,w):-\delta_1u+\delta_2v+\delta_3w=0\}$
3. $V=\{(u,v,w):\delta_1u+\delta_2v-\delta_3w=0\}$
4. $V=\{(u,v,w):\delta_1u+\delta_2v+\delta_3w=0\}$

Solution:

Consider this determinant,
\begin{vmatrix}
u & v & w  \\
x_1 &x_2&x_3 \\
y_1&y_2&y_3 \\
  \notag
\end{vmatrix}
If $\{u,v,w\}$ spans \(\mathbb{R}^3\) thus also spans \(x,y\) and thus the value of this determinant is zero.
The expression written in the first option is the value of this determinant
so, 1st option is correct.


3. Let $A=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$ .Let $f:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}$ be defined by $f(v,w)=w^TAv$,
Pick the correct statement from below:

1. There exists an eigenvector $v$ of $A$ such that $Av$ is perpendicular to $v$
2. The set $\{v\in \mathbb{R}^2 |f(u,v)=0\}$ is a nonzero subspace of $\mathbb{R}^2$
3. If $u,v\in \mathbb{R}^2$, are non-zero vectors such that $f(v,v)=0=f(w,w)$, then $v$ is a scala  multiple of $w$
4. For every $v\in \mathbb{R}^2$, there exists a nonzero $w\in \mathbb{R}^2$ such that $f(v,w)=0$

Solution: 

For every $(u,v)\in \mathbb{R}$ there will exists $(v,u)\in \mathbb{R}^2$ such that

$\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ $\begin{bmatrix}v\\u\end{bmatrix}=0$

So,

4th option is correct.

4. Let $A$ be a $n\times m$  matrix and $b$ be $n\times 1$ vector (with real entries), suppose the equation $Ax=b, x\in \mathbb{R}^m$ admits a unique solution, Then we can conclude that

1. $m\geq n$
2. $n\geq m$
3. $n=m$
4. $n>m$

Solution:Let $A_{n\times m} $ matrix has rank $r$ then there exists unique solution in two cases

• $n=m=r$
• $r=m<n$

Combining two cases, We get $n\geq m$

So, 2nd option is correct.

5.Let $V$ be the vactor space of all real polynomials of degree $\leq 10$. Let $Tp(x)=p'(x)$ for $p\in V$ be a linear transformation from $V$ to $V$. Consider the basis $\{1,x,x^2,\cdots x^{10}\}$ of $V$. Let $A$ be the matrix of $T$ with respect to this basis. Then 
Trace$A$=1
Det $A$=0
There is no $m\in \mathbb{N}$ Such that $A^m=0$
$A$ has a nonzero eigenvalue

Solution:
The matrix of above linear transformation is
$\begin{bmatrix}0&1&0&0&\cdots&0\\0&0&2&0&\cdots&0\\0&0&0&3&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\0&0&0&0&\cdots&0\end{bmatrix}$
This is an upper triangular matrix so, It's determinant is equal to zero.
So, 2nd option is correct.

6. Let $A$ be a $n\times n$ real symmetric non-singular matrix. Suppose there exists $x\in \mathbb{R}^n$ such that 
$x'Ax<0$
Then we can conclude that

1. $det(A)<0$
2. $B=-A$ is positive definite
3. $\exists y\in \mathbb{R}^n \quad y'A^{-1}y<0$
4. $\forall y\in \mathbb{R}^n \quad y'A^{-1}y<0$

Solution:

Option 1:
Consider,
 $A=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ Clearly $\det(A)>0$
But $\exists x=\begin{bmatrix}1&0\end{bmatrix}$ such that $ \begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}-1&0\\0&-1\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}<0$
So, 1st option is not correct.

Option 2:
We can not coclude that $A$ is negative definite since criteria for being negative definite is:
$\forall x\in \mathbb{R}^n$ such that
$x'Ax<0$
But here this condition is not given.

Option 3:
Take now $\;v:=Ax\;$ , then

$$v^tA^{-1}v=x^tA^tA^{-1}Ax=x^tA^tx=x^tAx<0$$

option 4: 
$A:=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\;\;\text{is a counterexample to 4th option  since}\;\; (0\;1)A\binom01>0$

$\text{but}\;\;(1\;0)A\binom10<0\;\;\text{and}\;\;A^{-1}=A$

So, 3rd option is correct.

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CSIR-NET Mathematical science Paper : June 2016

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Some Solved problems on zeros and poles.

Few Useful results regaring the poles and zeros in terms of quotients and product of two functions,

• Let \(f(z)=h(z)/g(z)\), where \(h\) and \(g\) are analytic in some open disk about \(z_0\), then \(f\) has a pole of order \(n\) at \(z_0\).
• Let \(f(z)=h(z)/g(z)\) and suppose \(h\) and \(g\) are analytic in some open disk about \(z_0\), with \(n>k\), then \(f\) has a pole of order \(n-k\) at \(z_0\).
• Let \(f\) has a pole of order \(m\) at \(z_0\) and let \(g\) has a pole of order \(n\) at \(z_0\). then \(fg\) has a pole of order \(m+n\) at \(z_0\)
• Given a function $f(z)$, we say that $f$ has a singularity (of a given type) at infinity if and only if $f(\frac1z)$ has a singularity (of said type) at $0$.

Here are some solved examples:

1. \(f(z)=\frac{1+4z^3}{sin^6z}\) 

Solution:

at \(z=n\pi\) denominator becomes zero. and numerator is not zero at any of these points.

so, \(f(z)\) has pole of order \(6\) at \(n\pi\)

2. \(f(z)=\frac{(z-\frac{3\pi}{2})^4}{\cos^7z} \)

Solution:

at \(\frac{3\pi}{2}\) numerator has zero of order 4. and at this point denominator has zero of order \(7\).

so, \(f(z)\) has a pole of order \(3\) at this point.
moreover, pole of order 7 at \(n\pi/2\).(other than \(3\pi/2\).)

3.  \(f(z)=\frac{1}{z^2 sin z}\) 

solution:

at \(z=0\), \(\frac{1}{z^2}\) has pole of order 2 and \(\frac{1}{\sin z}\) has a pole of order \(1\) at \(z=0\)

so, \(f(z)=\frac{1}{z^2 sin z}\) has a pole of order \(3\) at \(0\).
and simple pole at \(z=n\pi\).

4. \(f(z)=z^2-\frac{1}{z^2}\) 

Solution:

\(f(z)=z^2-\frac{1}{z^2}=\frac{z^4-1}{z^2}\)

at \(z=0\), \(f(z)\) has pole of order \(2\).

at \(z=1\) \(f(z)\) has zero of order \(1\).

5. \(f(z)=\frac{1}{(z^2+a^2)^2}\).

Solution:

pole at  \(z=\pm a i\) of order \(2\).

6. \(f(z)=\frac{\sin^2 z}{z^2}\) 

Solution:

numerator:

at \(z=0\) zero of order \(2\)

Denominator:

At \(z=0\), zero of order \(2\).

So, at \(z=0\), \(f(z)\), has no singularity.

and zero of order \(2\) at \((2n+1)\pi\).

7. \(f(z)=(z+1)\sin(\frac{1}{z-2})\)

Solution:

\(=(z+1)(\frac{1}{z-2})+\frac{1}{z-2}^3\times \frac{1}{3!}+\cdots\)

\(=\frac{z+1}{z-2}+\frac{z+1}{(z-2)^3}\frac{1}{3!}+\cdots\)

essential singularity at \(z=2\)

zero at \(z=-1\)

8. \(f(z)=\frac{e^{2z}}{(z-1)^4}\)

Solution:

numerator is not zero for any \(z\).

and denominator has zero of order 4 at \(1\).

so \(f(z)\) has pole of order at \(z=0\) of order \(4\).

9. \(f(z)=\cos^2(\frac{z}{2})\)

Solution:

zero at \(z=n\pi\) of order 2.

10. \(f(z)=(z^2+1)(e^z-1)\)

solution:

at \(z=0\)  zero of order \(1\).

at \(z=\pm i\) zero of order \(1\)

11. \(f(z)=(z^4-z^2-6)^3\)

Solution:

\(Z^4-Z^2-6=0 \Rightarrow z=3,-2\)

at \(z=-2,3\) zero of order \(3\).

12. \(f(z)=\frac{z^4}{\sin z}\) 

Solution:

Numerator: zero of order \(4\) at \(z=4\)

Denominator: zero of order \(1\) at \(n\pi\)

So, at \(z=0\) zero of order \(3\)

and at \(z=n\pi\) pole of order \(1\).

13. \(f(z)=\sin \frac{1}{z}\)

Solution:

essential singularity at \(z=0\)

14. \(f(z)=\frac{1-\cot z}{z}\)

Solution:

zero of order \(1\) at \(2\pi +\frac{\pi}{4}\)

and pole of order \(1\) at \(0\)

15. \(f(z)=cos^3 z\)

Solution: 

zero of order \(3\) at \((2n+1)\frac{\pi}{2}\)

16. \(f(z)= \frac{(\pi-2)(z^4-3z^2)}{\sin^2 z}\)

Solution:

At \(z=0\) numerator has zero of order \(3\)

and denominator has zero of order \(2\)

so, \(f(z)\) has zero of order 1 at this point.

At \(z=\pi)\)  numerator has zero of order \(1\)

and denominator has zero of order \(2\)

so, \(f(z)\) has pole of order 1 at this point.

now let look \(1/f(z)\)

\(1/f(z)=\frac{\sin^2z}{(\pi-z)(z^4-z^2)}\)
and this function has essential singularity at \(z=0\).

so, \(f(z)\) has essential singularity at \(z=\infty\).

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Relation and Function - NCERT Exemplar Problems and Solution (Short Answer Type)

Short Answer Type NCERT Exemplar Problems and Solution

Question – 1 Let$ A = (a, b, c)$ and the relation $R$  be defined on $A$  as follows:
$R = ((a, a), (b, c), (a, b)).$
Then, write minimum number of ordered pairs to be added in $R $ to make $ R$ reflexive and transitive.

Solution:

In order to make R reflexive,$ (b, b) $ and $(c, c) $ will be added to $R$

And in order to make $ R$  transitive,$ (a, c)$ will be added to $R.$

Therefore, The minimum number of order pair to be added to $R$ will be $(b, b), (c, c) $ and $ (a, c) $ ♦

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Long Answer Type NCERT Exemplar Problems and Solution Part 1

Question – 16: If $A = (1, 2, 3, 4)$, define relations on $A$ which have properties of being

(a) Reflexive, transitive but not symmetric

(b)  Symmetric but neither reflexive nor transitive.

(c)  Reflexive, symmetric and transitive.

Solution: Let $R_1 $= { $(1,1),(2,2),(1,2)$}

Thus, it is clear that $(1,1)\in R_1$ and $(2,2)\in R_1$

Thus $R_1$ is reflexive.

Again, $(1,2)\in R_1$ $but\hspace{5pt} (2,1)\notin R_1 $

Thus it is not symmetric.Now,  $(2,1) \in R_1, $ and $(1,1)\in R_1 $ $ \Rightarrow (2,1)\in R_1$ Thus $R_1$ is reflexive and transitive but not symmetric.

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CSIR Part A: Solved: December 2015 (Mathematics)

Question 1:

A circle drawn in the $x-y$ coordinate plane passes through the origin and has chords of lengths $8$ units and $7$ units on the $x$ and $y$ axes, respectively. The coordinates of its centre are

1.    (8, 7)                                                                                                 2.     (-8, 7)
3.   (-4, 3.5)                                                                                            4.      (4, 3.5)

Solution:

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CSIR-NET Mathematical Science and Life Science (Part A Solved):2015 June

1. A pyramid shaped toy is made by tightly placing cubic blocks of $1\times 1 \times 1$ $cm^3$. The base of the toy is a square  $4\times 4$ $Cm^2$. The width of each step is 0.5 Cm . how many blocks are required to make the toy?

Solution:

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2015 (December) CSIR Part A (Solution): Engineering Science, Chemical Science and physical Science.

1.Each of the following pairs of words hides a number, based on which you can arrange them in ascending order. Pick the correct answer:

I. Cloth reel

J. Silent Wonder

K. Good tone

L . Bronze rod

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Part A: CSIR - NET Solution (Bio Group):Dec 2015

1. A shopkeeper purchased a product for Rs.$100$ and sells it making a profit of $10$%. The customer resells it to the same shopkeeper incurring a loss of $10$%. In these dealings the shopkeeper makes
1. no profit, no loss

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