AKU Previous year questions from Matrices:

 [AKU- Year 2020]

Q1. If $A=\begin{bmatrix}2 &0&0\\0&2&0\\0&0&2\end{bmatrix}~ and~ B=\begin{bmatrix}1&2&3\\0&1&3\\0&0&2\end{bmatrix}$                                        

Then the determinant of $AB$ has the value

a) 4            b) 8        c) 16        d) 32

Q2. The sum and product of the eigenvalues of 

$\begin{bmatrix}2&2&1\\1&3&1\\1&2&2\end{bmatrix}$ are                               

a) 7 and 5            b) 5 and 7        c) 12 and 3        d) 3 and 12

[AKU- Year 2018]

Q3. If the nullity of the matrix $\begin{bmatrix}k&1&2\\1&-1&-2\\1&1&4 \end{bmatrix}$ is 1. Then the value of $k$ is                                                                                                                                                  

a) -1                    b) 0            c)  1        d) 2                                                       

Q4.  Let $A=\begin{bmatrix}3&0&0\\0&6&2\\0&2&6\end{bmatrix}$ and let $\lambda_1 \geq \lambda_2 \geq \lambda_3$ be the eigenvalues of $A$. Then the triple $(\lambda_1, \lambda_2, \lambda_3)$equals

a) (9, 4, 2)        b) (8,4,3)    c) (9,3,3)    d)(7,5,3)                         

Q5. If $A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$ then $A^{50}$ is                                                                             

a) $\begin{bmatrix}1&0&0\\50&0&0\\50&0&1\end{bmatrix}$ b) $\begin{bmatrix}1&0&0\\48&0&0\\48&0&1\end{bmatrix}$ c) $\begin{bmatrix}1&0&0\\25&1&0\\25&0&1\end{bmatrix}$ d)$\begin{bmatrix}1&0&0\\24&1&0\\24&0&1\end{bmatrix}$ 

[AKU- Year 2019]

Q6. If the eigen values of the given matrix                                

$\begin{bmatrix}-2&2&-3\\2&1&-6\\-1&-2&0\end{bmatrix}$

is 3. then the eigen value of $adj (A)$ will be

a) $-\frac{1}{3}$      b) $-\frac{1}{5}$        c)$-\frac{1}{15}$         d) -3

Q7. Write down the quadratic forms corresponding to the given matrix

$\begin{bmatrix}2&4&5\\4&3&1\\5&1&1\end{bmatrix}$     

Subjective Questions:    

AKU- YEAR 2020   

Q1. Determine the value of $p$ such that the rank of 

$\begin{bmatrix}1&1&-1&0\\4&4&-3&1\\p&2&2&2\\9&9&p&3\end{bmatrix}$ is 3

Solution: Click here for solution. 

Q2. Use Gauss-Jordan method to find the inverse of the matrix

$\begin{bmatrix}2&3&4\\4&3&1\\1&2&4\end{bmatrix}$ 

Solution: Click here for solution. 

Q3. Find the non-singular matrices $P$ and $Q$ such that

$\begin{bmatrix}1&2&3&4\\2&1&4&3\\3&0&5&-10\end{bmatrix}$

is reduced to normal form. Also find its rank.

Solution: Click here for solution. 

Q4. Solve the given equations by Cramer's rule

$x+y+z=4$

$x-y+z=0$

$2x+y+z=5$

Solution: Click here for solution.

 Q5. Verify Cayley-Hamilton theorem for the matrix

$\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$ and find the inverse. 

Solution: Click here for solution.

Q6. Find the eigen vectors of the matrix

$\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$

Hence deduce 

$6x^2+3y^2+3z^2-2yz+4zx-4xy$

to a 'sum of squares'. Also write nature of the matrix.

Solution: Click here for solution. 

Q7. Find the eigen values and eigen vectors of the matrix

$\begin{bmatrix}-2&2&-3\\2&1&-6\\-1&-2&0\end{bmatrix}$

Solution: Click here for solution.

Q8. For what values of $k$, the equations

$x+y+z=1,~2x+y+4z=k,~4x+y+10z=k^2$

Have solution? Solve them completely in each case.

Solution: Click here for solution.

Q9. Reduce the matrix

$\begin{bmatrix}-1&2&-2\\1&2&1\\-1&-1&0\end{bmatrix}$ to the diagonal form.

 Solution: Click here for solution.

Q10. Find the rank of the matrix

$\begin{bmatrix}2&3&-1&-1\\1&-1&-2&-4\\3&1&3&-2\\6&3&0&-7\end{bmatrix}$

Solution: Click here for solution.

Q11. Find the characteristic equation of the matrix

$A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$

and hence find the matrix represented by

$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$

Solution: Click here for solution.

Q12. State and prove Cayley-Hamilton theorem.

Solution: Click here for solution.

Q13. Reduce the quadratic form $3x^2+5y^2+3z^2-2xy-2yz+2zx$ to canonical forms.

Solution: Click here for solution.

AKU- YEAR 2018  

 Q14. Let $T:R^3\rightarrow R^4$ be a linear transformation defined by

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x+y\\z+y\\x+z\\x+y+z\end{bmatrix}$

Find the matrix representation of $T$ with respect to the ordered basis

$x=\lbrace \begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}\rbrace$ in $R^3$ and $y=\lbrace \begin{bmatrix}0\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\1\\1\end{bmatrix},\begin{bmatrix}1\\1\\0\\1\end{bmatrix},\begin{bmatrix}1\\1\\1\\0\end{bmatrix}\rbrace $ in $R^4$

Solution: Click here for solution.

 

Central Tendency

Central Tendency:

Data can be classified in various forms. One way to distinguish between data is in terms of grouped and ungrouped data.

What is ungrouped data?

When the data has not been placed in any categories and no aggregation/summarization has taken placed on the data then it is known as ungrouped data. Ungrouped data is also known as raw data.

Height of students:

 (171,161,155,155,183,191,185,170,172,177,183,190,139,149,150,

150,152,158,159,174,178,179,190,170,143,165,167,187,169,182,

163,149,174,174,177,181,170,182,170,145,143):

This is raw/ungrouped data.

When raw data have been grouped in different classes then it is said to be grouped data.

Height of students:

(139, 143, 143, 145, 149, 149,150,150,152, 155,155, 158,159, 161, 163, 165,167, 169, 170, 170, 170 170, 171, 172, 174, 174,174, 177, 177, 178,179, 181, 182, 182, 183, 183, 185, 187, 190, 190, 191)

Before we study more about grouped and ungrouped data it is important to understand what do we mean by “Central Tendencies”?

 

Measures of central tendency

These are statistical constants which give us an idea about the concentration of the values in the central part of the distribution. The various measures of central tendency are:

1. Arithmetic Mean
2. Median
3. Mode
4. Gerometric Mean
5. Harmonic Mean

• Arithmetic Mean

    Arithmetic mean of a set of observations is their sum divided by

 the number of observations, E.g., the Arithmetic mean $\bar{x}$ of $n$ observations $x_1, x_2, x_3,........, x_n$ is given by

                  $\bar{x}=\frac{x_1+x_2+x_3+....+x_n}{n}=\frac{1}{n}\sum_{i=1}^n x_i$ (Ungrouped data) 

        In case of frequency distribution $\ x_i, f_i, i=1,2,3,......, n$ where $\ f_i$ is the frequency of the variable $\ x_i$

$\bar{x}$=$\frac{f_1x_1+f_2x_2+f_3x_3+.......+f_nx_n}{f_1+f_2+f_3+.......+f_nx_n}$=$\frac{\sum_{i=1}^n f_ix_i}{\sum_{i=1}^n f_i}$ (Grouped Data with frequency Distribution)  

       In case of continuous frequency distribution, $x$ is taken as the mid value of the corresponding class.

 

Example: Find the arithmetic mean of the following distribution

        (a) x: 150,200,300,650,250,180,400,500,550,220

Solution (a): Here the data is group data. So the arithmetic mean is

           $\bar{x}$ = $\frac{150+200+300+650+250+180+400+500+550+220}{10}$=340

Solution (b) :  Here the given data are grouped data with discrete frequency distribution. So the arithmetic mean is

                  $\bar{x}$=$\frac{1\times5+2\times9+3\times12+4\times17+5\times14+6\times10+7\times6}{5+9+12+17+14+10+6}$=4.09

Example: Find the arithmetic mean of the following distribution

                            If the values of $x$ or $f$ are large, the calculation of mean is quite time consuming and tedious. The arithmetic is required to a great extent by taken deviations of the given values from any arbitrary point "A".

                      

                      $d_i$ = $x_i - A$ ⇒$\sum_{i=1}^n d_i$ = $\sum_{i=1}^n x_i - A_n$

 

            Now , 

                         $\frac{\sum_{i=1}^n d_i}{n}$ = $\frac{\sum_{i=1}^n x_i}{n}$ - $A$

                     

                         $\bar{x}$ = $A$ + $\frac{\sum_{i=1}^n d_i}{n}$         (for ungrouped data)  

 

         

                            $\bar{x}$ = $A$+$\frac{1}{N}{\sum f_id_i}$     where $N$=$\sum f_i$

 

 

                            $d_i$ = $\frac{x_i-A}{h}$

                        $\bar{x}$= $A$+$\frac{h}{N}{\sum f_id_i}$      where, $N$= $\sum f_i$

 

 Example : Calculate the mean for the following frequency distribution

 

Solution :

  

                            

Example : The average salary of male employees in a firm was Rs. 5,200 and that of female was Rs. 4,200. The mean salary of all the employees was Rs. 5000. Find the percentage of male and female employees.   

 

  

Bi-variate Distributions

Bivariate Distributions

Joint Distribution function: Let X and Y be two random variables defined on the same probability space $(\Omega, \tilde{A}, P[.])$. Then (X, Y) is called a two-dimensional random variable.  The joint cumulative distribution function or joint distribution function of X and Y, denoted by $F_{X,Y}(x,y)$, is defined as

$$F_{X,Y}(x,y)= P[X \leq x, Y \leq y]~ \forall x, y \in R $$

It may be observed that the joint distribution function is a function of two variables and its domain is the xy-plane. Sometimes we write $F_{X,Y}(x,y)$ as $F(x,y)$.

Properties of Joint Distribution function:

1.    If $x_1<x_2$ and $y_1<y_2$ then (rectangle rule)

$$ P[x_1 <X \leq x_2, y_1<Y\leq y_2]=F(x_2,y_2)-F(x_2,y_1)-F(x_1,y_2)+F(x_1, y_1) \geq 0$$

2.    (a) $F(-\infty, y)=\lim_{x \to -\infty} F(x,y)=0 ~\forall y \in R$

       (b) $F(x, -\infty)=\lim_{y \to -\infty}F(x,y)=0 ~\forall x \in R$

       (c) $F(x, -\infty)= \lim_{x\to \infty, y\to \infty}F(x,y)=1$

3. $F(x, y)$ is right continuous in each argument i.e.

$$\lim_{h\to 0_+} F(x+h, y) = lim_{h \to 0_+} F(x, y+h)=F(x, y)$$

Remark: Any function of two variables which fails to satisfy one of the above three conditions is not a joint distribution function.

Example 1. Show that the bivariate function: 

$F(x, y) = \left\{\begin{array}{ll} e^{-(x+y) & \quad x>0, y>0 \\ 0 & \quad otherwise\end{array}\right.$

is not a joint distribution function.

Solution: $x^2$

Multinomial, Exponential and Gamma distributions

Multinomial, Exponential and Gamma distributions  

Multinomial Distribution: This distribution can be regarded as a generalization of Binomial distribution.

When there are more than two exclusive outcomes of a trial, the observation leads to Multinomial distribution.

Suppose $E_1, E_2,......, E_k$ are mutually exclusive and exhaustive outcomes of a trial with respective probabilities $p_1, p_2, .....,p_k$.  The probability that, $E_1$ occurs $x_1$ times, $E_2$ occurs $x_2$ times,………, $E_k$ occurs $x_k$ times in $n$ independent observations, is given by

$$P[x_1, x_2,......, x_k]=C. p_1^{x_1} p_2^{x_2}....... p_k^{x_k}, ~ where, \sum x_i =n$$

and $C$ is the number of permutation of the events $E_1, E_2,.....,E_k$ with $C=\frac{n!}{x_1!x_2!.....x_k!}$

Therefore, $P[x_1,x_2,....,x_k]=\frac{n!}{\prod_{i=1}^k x_i!} \prod_{i=1}^k p_i^{x_i}, 0\leq x_i \leq n$

Also $(p_1+p_2+.......+p_k)^n=1$ as $\sum p_i =1$.

Example 1. There are given a bag of marbles. Inside the bag, there are 5 red marbles, 4 white marbles and 3 blue marbles. Calculate the probability that with 6 trials, we have to choose 3 marbles that are red, 1 marble that is white and 2 marbles that are blue, replacing each marble after it is chosen.

Solution: Here total number of marbles are 12. Therefore probabilities of selecting red, white and blue marbles are $p_r=\frac{5}{12}, p_w=\frac{4}{12}$ and $p_b=\frac{3}{12}$ respectively.

Let $E_r, E_w$ and $E_b$ denote the events of choosing 3 red, 1 white and 2 blue marbles, then the number of marbles to be chose $x_r=3, x_w=1$ and $x_b=2$ and number of trials $n=6=x_r+x_w+x_b$.

Now the number of permutation of the events $E_r, E_w$ and $E_b$ is $C=\frac{6!}{3!1!2!}$

Therefore, probability that $E_r, E_w$ and $E_b$  occur 3 times, 1 time and 2 times, is

$P[x_r=3, x_w=1, x_b=2]=C. p_r^{x_r} p_w^{x_w} p_b^{x_b}$

$P[x_r=3, x_w=1, x_b=2]=\frac{6!}{3!1!2!} \left( \frac{5}{12}\right)^3 \left(\frac{1}{12}\right)^1 \left(\frac{3}{12}\right)^2$

$$P[x_r=3, x_w=1, x_b=2]=0.0899$$

Exercise: We are randomly drawing cards from an ordinary deck of cards. Every-time we pick one, we place it back in the deck. We do this 5 times. What is the probability of drawing 1 heart, 1 space, 1 club and 2 diamonds? Ans. 0.0586 

Exponential and Gamma distributions

Exponential distribution: A random variable X is said to have an exponential distribution with parameter $\lambda>0$, if its density function is given by $f_X(x)=\lambda e^{-\lambda x}$ for all $x\geq 0$ and $\lambda>0$

We write it as $X \sim Expo(\lambda)$

Gamma distribution: A random variable X is said to have a gamma distribution if its density function is given by $f_X(x)=\frac{\lambda}{\Gamma(r)} (\lambda x)^{r-1}e^{-\lambda x}$ for all $x\geq 0$,

Where $r>0$ and $\lambda >0$ are called the parameters of the gamma distribution.

We write it as $X \sim gam(\lambda; r)$ or $X\sim G(\lambda ; r)$

Remark 1. $\Gamma(r) $ is called gamma function and is defined as $\Gamma(r)=\int_0^{\infty} x^{r-1}e^{-x}dx$

It is easy to verify that $\Gamma(1)=1, \Gamma(a+1)=a\Gamma(a), \Gamma(n)=(n-1)!, n$ is a positive integer. 

Remark 2. Taking $r=1$, we see that gamma density function becomes exponential density function.

Theorem: If X has an exponential distribution, then  

$$E[X]=\frac{1}{\lambda}, var[X]=\frac{1}{\lambda^2}~ and~ M_X(t)=\frac{\lambda}{\lambda-t} ~for~ t< \lambda$$ 

Proof: We have $E[X]=\int_0^{\infty} xf_X(x) dx = \int_0^{\infty}x \lambda e^{-\lambda x}dx, \lambda>0$

Integrating by parts, we get

$$E[X]=\lambda \left| x \left(\frac{e^{-\lambda x}}{-\lambda}\right)\right|_0^{\infty}+\int_0^{\infty} 1. e^{-\lambda x} dx = 0-\frac{1}{\lambda}\left| e^{-\lambda x}\right|_0^{\infty}=-\frac{1}{\lambda}(0-1)$$

$$E[x]=\frac{1}{\lambda} ....... (1)$$

Now $E[x^2]=\int_0^{\infty} x^2 f_X(x) dx =\int_0^{\infty} x^2 \lambda e^{-\lambda x} dx$

Integrating by parts, we get

$$E[X^2]=\lambda \left| x^2 \left( \frac{e^{-\lambda x}}{-\lambda}\right)\right|_0^{\infty}+ 2 \int_0^{\infty} x. e^{-\lambda x} dx=0+0+\frac{2}{\lambda} \int_0^{\infty} e^{-\lambda x}dx$$

$$E[X^2]=\frac{2}{\lambda}.\frac{1}{\lambda}=\frac{2}{\lambda^2}$$

Therefore, $Var[X]=E[X^2]-(E[X])^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}$

The m.g.f. of X is given by

$$ M_X(t) = E[e^{tX}]=\int_0^{\infty} e^{tx}f_X(x) dx=\int_0^{\infty} e^{tx}\left(\lambda e^{-\lambda x}\right)dx $$

$$ M_X(t)=\lambda \int_0^{\infty} e^{-(\lambda-t)x}dx,~~~ \lambda >t$$

$$ M_X(t)=- \frac{\lambda}{\lambda-t}\left| e^{-(\lambda-t)x}\right|_0^{\infty}=-\frac{\lambda}{\lambda-t}(0-1)$$

Hence, $$ M_X(t)=\frac{\lambda}{\lambda - t}, ~ for ~\lambda > t$$.

Theorem: If X has gamma distribution with parameters $r$ and $\lambda$, then

$$E[X]=\frac{r}{\lambda},~ var[X]=\frac{r}{\lambda^2}~ and~ M_X(t)=\left(\frac{\lambda}{\lambda-t}\right)^r~ for ~t<\lambda$$

Theorem: Show that the sum of independent gamma variates is also a gamma variate

Hint: Let $x_i \sim G(\lambda; r_i)$ for $i=1, 2, ......., n;$ 

Example 1. If X has exponential distribution with mean 2, find $P[X<1|X>2]$

Solution: We are given that $\frac{1}{\lambda}=2 \Rightarrow \lambda=\frac{1}{2}$

Now $P[X<1|X<2]=\frac{P[(X<1)\cap (X<2)]}{P[X<2]}=\frac{P[X<1]}{P[X<2]}=\frac{\int_0^1 \lambda e^{-\lambda x}dx}{\int_0^2 \lambda e^{-\lambda x}dx}$

$P[X<1|X<2]=\frac{-\frac{1}{\lambda}|e^{-\lambda x}|_0^{\infty}}{\frac{1}{\lambda}|e^{-\lambda x}|_0^2}= \frac{1-e^{-\lambda}}{1-e^{-2\lambda}}=\frac{1}{1+e^{-\lambda}}$

Hence, $P[X<1|X<2]=\frac{1}{1+e^{-\frac{1}{2}}}$

Exercises:

1. If X has exponential distribution with $P[X\leq 1]=P[X>1],$ then find $var[x]

2.  Find the median of the exponential distribution. 

3. If $X\sim Expo (\lambda)$ find the value $k$ such that $\frac{P[X>k]}{P[X\leq k]}=a.$