GATE Breaker

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & i & i \\ 0 & 0 & 0 & -i \end{bmatrix}$, then find the value of $A^4$ using Cayley-Hamilton theorem.  (GATE 1993)  Answer: $A^4 = I$.  Explanation:  Since the given matrix $A$ is an upper triangular matrix, the eigenvalues of $A$ are the main diagonal values.  Therefore, the eigenvalues of $A$ are $1$, $-1$, $i$, $-i$.  Hence, the characteristic equation of of $A$ is  $(\lambda - 1)(\lambda + 1)(\lambda - i)(\lambda + i) = 0$   $\Rightarrow  (\lambda^2 - 1)(\lambda^2 + 1) = 0$  $\Rightarrow  (\lambda^4 - 1) = 0$  Therefore, by Calley-Hamilton theorem, we have $A^4 - I = 0$, i.e., $A^4 = I$. 

Let $P(E)$ denote probability of an event $E$. Given $P(A) = 1$, $P(B) = \dfrac{1}{2}$. The values of $P(A|B)$ and $P(B|A)$ respectively are  (A) $\dfrac{1}{4}$, $\dfrac{1}{2}$  (B) $\dfrac{1}{2}$, $\dfrac{1}{4}$  (C) $\dfrac{1}{2}$, 1  (D) 1, $\dfrac{1}{2}$  (GATE 2003)  Answer: (D) 1, $\dfrac{1}{2}$  Explanation:  Since, $P(A) = 1$ and $P(B) = \dfrac{1}{2}$, hence, $P(A \cap B) = P(B) = \dfrac{1}{2}$.  Now, $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$,  and $P(B|A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}$. 

Four fair coins are tossed simultaneously. The probability that at least one heads and at least one tails turn up is?  (A) $\dfrac{1}{16}$,     (B)  $\dfrac{1}{8}$,     (C)  $\dfrac{7}{8}$,     (D)  $\dfrac{15}{16}$.  (GATE 2002)  Answer: (C) $\dfrac{7}{8}$  Explanation:  Probability of getting head, $p = \dfrac{1}{2}$.  Probability of getting tail, $q = \dfrac{1}{2}$ .  Therefore, the probability of getting at least one head and one tail is  $= {}^{4}\textrm{C}_{1} p^{1} q^{3} + {}^{4}\textrm{C}_{2} p^{2} q^{2} + {}^{4}\textrm{C}_{3} p^{3} q^{1}$, from binomial distribution  $= {}^{4}\textrm{C}_{1}\left(\dfrac{1}{2}\right)^{1}\left(\dfrac{1}{2}\right)^{3} + {}^{4}\textrm{C}_{2}\left(\dfrac{1}{2}\right)^{2}\left(\dfrac{1}{2}\right)^{2} + {}^{4}\textrm{C}_{3}\left(\dfrac{1}{2}\right)^{3}\left(\dfrac{1}{2}\right)^{1}$  $= \dfrac{7}{8}$ 

Seven car accidents occurred in a week, what is the probability that they all occurred on the same day?  (A) $\dfrac{1}{7^7}$,     (B) $\dfrac{1}{7^6}$,     (C)  $\dfrac{1}{2^7}$,     (D) $\dfrac{7}{2^7}$  (GATE 2001)  Answer:  (B) $\dfrac{1}{7^6}$  Explanation:  The probability that all accidents occurred on Monday $= \dfrac{1}{7^7}$  The probability that all accidents occurred on Tuesday $= \dfrac{1}{7^7}$  $\cdots$      $\cdots$      $\cdots$      $\cdots$      $\cdots$      $\cdots$ $\cdots$    $\cdots$    $\cdots$    $\cdots$    $\cdots$    $\cdots$ The probability that all accidents occurred on Sunday $= \dfrac{1}{7^7}$  Therefore, the probability that they all occurred on the same day is  $7 \times \dfrac{1}{7^7} = \dfrac{1}{7^6}$. 

In a manufacturing plant, the probability of making a defective bolt is 0.1. The mean and standard deviation of defective bolts in a total of 900 bolts are respectively  (A) 90 and 9,     (B) 9 and 90,     (C) 81 and 9,     (D) 9 and 81.  (GATE 2000)  Answer:  (A) 90 and 9  Explanation:  Let $X$ be a random variable that denotes the number of defective bolts.  Here, $X$ follows a binomial distribution with the parameters $p = 0.1$ and $n = 900$.  Therefore, mean $= E(X) = n \times p = 0.1 \times 900 = 90$, and  standard deviation $= \sqrt{n \times p \times (1 -p)} = \sqrt{900 \times 0.1 \times (1 - 0.1)} = 9$. 

$E_1$ and $E_2$ are events in a probability space satisfying the following constraints: $P(E_1) = P(E_2)$, $P(E_1 \cup E_2) = 1$.  If $E_1$ and $E_2$ are independent then $P(E_1) = $  (A) 0,     (B) $\dfrac{1}{4}$,     (C) $\dfrac{1}{2}$,     (D) 1  (GATE 2000)  Answer: (D) 1   Explanation:  Given, $P(E_1) = P(E_2)$, $P(E_1 \cup E_2) = 1$, and $E_1$ and $E_2$ are independent events.  Hence, $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$.  Now, $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1) \times P(E_2)$  $\Rightarrow  1 = P(E_2) + P(E_2) - P(E_2) \times P(E_2)$  $\Rightarrow  \{P(E_2)\}^2 - 2 \times P(E_2) + 1 = 0$  $\Rightarrow  (P(E_2) - 1)^2 = 0$  $\Rightarrow  P(E_2) = 1$. 

The probability that two friends share the same birth-month is (A) $\dfrac{1}{6}$,      (B) $\dfrac{1}{12}$,     (C) $\dfrac{1}{144}$,     (D) $\dfrac{1}{24}$  (GATE 1998)  Answer:  (B) $\dfrac{1}{12}$  Explanation:  Sample space, $S = $  $\{(Jan, Jan), (Jan, Feb), \ldots, (Jan, Dec), $ $\cdots$     $\cdots$      $\cdots$    $\cdots$      $\cdots$    $\cdots$      $\cdots, $ $(Dec, Jan), (Dec, Feb), \ldots, (Dec, Dec)\}$.  No. of sample points, $n(S) = 12 \times 12 = 144$.  Favorable events, $E = \{(Jan, Jan), (Feb, Feb), \ldots, (Dec, Dec)\}$.  No. of event points, $n(E) = 12$.  Therefore, the required probability that the two friends share the same birth-month is  $= \dfrac{n(E)}{n(S)} = \dfrac{12}{144} = \dfrac{1}{12}$. 

A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is  (A) $\dfrac{1}{6}$, (B) $\dfrac{3}{8}$, (C) $\dfrac{10}{8}$, (D) $\dfrac{1}{2}$.  (GATE 1998)  Answer: (B) $\dfrac{3}{8}$  Explanation:  Probability of getting an odd number when a die is rolled = $\dfrac{3}{6} = \dfrac{1}{2}$.  Hence, the probability of 'success', $p = \dfrac{1}{2}$.   No. of trials, $n = 3$.  Therefore, the required probability is  $= ^3C_1 \times \dfrac{1}{2} \times \left(\dfrac{1}{2}\right)^{3-1}$, from the Binomial distribution with parameters $n = 3$ and $p = \dfrac{1}{2}$  $= \dfrac{3}{8}$. 

The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow?  (A) 0.3,      (B) 0.25,     (C) 0.35,     (D) 0.4  (GATE 1997)  Answer:  (D) 0.4  Explanation:  Let $E_1$ be the event of raining today, $E_2$ be the event of raining tomorrow.  According to the given problem, $P(E_1) = 0.5$, $P(E_2) = 0.6$, and $P(E_1 \cup E_2) = 0.7$.  Therefore, the required probability is  $= P(E_1 \cap E_2) $ $= P(E_1) + P(E_2) - P(E_1 \cup E_2)$ $= 0.5 + 0.6 - 0.7 = 0.4$.  

The probability that a number selected at random between100 and 999 (both inclusive) will not contain the digit 7 is  (A) $\frac{16}{9}$,  (B) $\left(\frac{9}{10}\right)^3$,  (C) $\frac{27}{75}$,  (D) $\frac{18}{25}$.  (GATE 1995) Answer: (D) $\frac{18}{25}$.  Explanation:  All the numbers between 100 and 999 (both inclusive) have three digits.  No. of ways of choosing the left-most digit from 1 to 9 excluding 7 = 8.  No. of ways of choosing the middle digit from 0 to 9 excluding 7 = 9.  No. of ways of choosing the right-most digit from 0 to 9 excluding 7 = 9.  So, the total number of ways of selecting a number between 100 and 999 (both inclusive) that will not contain the digit 7 is = $8 \times 9 \times 9$.  Again, the total number of ways of selecting any number between 100 and 999 (both inclusive) is = $^{900}C_1 = 900$.  Therefore, the required probability is = $\frac{8 \times 9 \times 9}{900} = \frac{18}{25}...

How many substrings of different lengths (non-zero) can be found formed from a character string of length $n$? (A) $n$,  (B) $n^2$,  (C) $\frac{n(n - 1)}{2}$,  (D) $\frac{n(n + 1)}{2}$.  (GATE 1994)  Answer: (D) $\frac{n(n + 1)}{2}$.   Explanation:  The given string has length $n$.  No. of substrings of length $n = 1$  No. of substrings of length $n - 1 = 2$  $\cdots \cdots \cdots \cdots \cdots \cdots$  $\cdots \cdots \cdots \cdots \cdots \cdots$  No. of substrings of length $1 = n$  Therefore, the total number of substrings  $= 1 + 2 + \cdots + n$  $= \frac{n(n + 1)}{2}$. 

We need to choose a team of 11 from a pool of 15 players and also select a captain. Find the number of different ways this can be done.  (A) $\binom{15}{11}$,  (B) $11 \times \binom{15}{11}$,  (C) $15 \times 14 \times \cdots \times 5$,  (D) $(15 \times 14 \times \cdots \times 5) \times 11$.  Answer: (B) $11 \times \binom{15}{11}$  Explanation:  Number of ways of selecting a captain from 15 players  = $\binom{15}{1}$.  Remaining players in the pool $= 15 - 1  = 14$, and the remaining number of players to be chosen $= 11 - 1 = 10$.  Number of ways of selecting 10 players from a pool of 14 players = $\binom{14}{10}$.  Therefore, the total number of different ways of selecting 11 players from a pool of 15 players and also select a captain is  $= \binom{15}{1} \times \binom{14}{10}$  $= 15 \times \frac{14 \times 13 \times 12 \times 11 \times 10!}{4! \times 10!}$  $= \frac{15 \times 14 \times 13 \times 12 \times 11!}{4...

Suppose that the eigenvalues of matrix $A$ are 1, 2, 4. Find the determinant of $(A^{-1})^T$.  Answer: 0.125  Explanation:  The eigenvalues of matrix $A$ are 1, 2, 4.  Therefore, determinant of $A$, $|A|$ = product of the  eigenvalues of $A$ = 8.  Now, $|A^{-1}| = \frac{1}{|A|} = \frac{1}{8}$.  Hence, $|(A^{-1})^T| = |A^{-1}| = \frac{1}{8} = 0.125$. 

Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2 + \sqrt{-1})$ and $3$. Find the determinant of $P$.  (GATE 2016)  Answer: 15  Explanation:  Here, $P$ is a $3 \times 3$ real matrix.  The given eigenvalues are $2 + i$ and $3$.  Since $P$ is a real matrix, the third eigenvalue must be $2 - i$.  We know that the determinant of a real matrix is equal to the product of its eigenvalues.  Therefore, the determinant of $P$ = product of eigenvalues = $(2 + i) \times 3 \times (2 - i)$ = $15$. 

Find the determinant of the matrix \[\begin{bmatrix} 6 & -8 & 1 & 1\\ 0 & 2 & 4 & 6\\ 0 & 0 & 4 & 8\\ 0 & 0 & 0 & -1 \end{bmatrix}\].  (A) 11, (B) -48, (C) 0, (D) -24 (GATE 1997) Answer: (B) -48  Explanation:  The given matrix is an upper triangular matrix.  We know that the determinant of an upper triangular matrix is the product of the main diagonal elements.  Therefore, the determinant of the given matrix is $= 6 \times 2 \times 4 \times (-1) = -48$.   Previous Post Next Post

How many undirected graphs (not necessarily connected) can be constructed out of a given set $V = \{v_1, v_2, \ldots, v_n\}$ of $n$ vertices? (GATE 2001) (A) $\frac{n(n-1)}{2}$, (B) $2^n$, (C) $n!$, (D) $2^{\frac{n(n-1)}{2}}$  Answer: (D) $2^{\frac{n(n-1)}{2}}$  Explanation:  Here, number of vertices = $n$.  Therefore, $^nC_2 = \frac{n(n-1)}{2}$ number of edges can be formed.  Now, each subset of these edges can form an undirected graph (not necessarily connected).  Therefore, total $2^{\frac{n(n-1)}{2}}$ number of graphs can be formed.  Next Post