Since, the given matrix is an upper triangular one, all eigen values are A. And hence A - λI = A.
So the question as has
(A - λI)X = 0
AX = 0

What x₁, x₂, x₃ are suitable?
Which means:
x₁ times column 1 + x₂ times column 2 + x₃ times column 3 = zero vector
Since α is not equal to zero, so x₃ must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.

Note: Out of syllabus.

In this DE, degree is 1 then this represent linear equation.

Note: Out of syllabus.

Note: Out of syllabus.

Note: Out of syllabus.

Note: Out of syllabus.

Since the given expression is in 0/0 form, so we can apply L-Hospital rule.

Let λ be eigen value, then characteristic equation will be
(1-λ) (-1-λ) (i-λ) (-i-λ)
= (λ²-1) (λ²+1)
= λ⁴-1
Characteristic equation is λ⁴-1 = 0.
According to Cayley-Hamilton theorem, every matrix satisfies its own characteristic equation, so
A⁴ = I

So finally, we can write

Circuit behaves as shift register and mod-6 counter. Note that this is the Johnson counter which is the application of shift register. And Johnson counter is mod-2N counter.

Total bit per character
= 8 bit data + 2 stop bit + 1 start bit
= 11 bits
No. of characters = 2400/11 = 218.18
Since, it is asked for transmitted characters we take floor and answer is 218.

(a) 1*2² + 1*2¹ + 0*2⁰ + 1*2^-1 + 0*2^-2 + 1*2^-3
= 4 + 2 + 0 + 0.5 + 0 + 0.125
= 6.625
(b) 1118 mod 16 = E, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4
Writing the mods result in reverse order gives (45E)_H.

Input will consist of 8 bit (two 4-bit numbers) = 2⁸ address.
Output will be of 8 bits.
So memory will be of 2⁸ × 8.
So, M = 256, N = 8.

RPM = 2400
So, in DOS, the disk rotates 2400 times.
Average latency is the time for half a rotation
= 0.5×60/2400 s
= 12.5 ms
In one full rotation, entire data in a track can be transferred. Track storage capacity = 62500 bits
So, disk transfer rate
= 62500 × 2400/60
= 2.5 × 10⁶ bps
So,
P = 12.5, Q = 2.5×10⁶

Time to service an interrupt
= saving state of CPU + ISR execution + restoring of CPU state
= (80 + 10 + 10) × 10^-6
= 100 μs
For every 1ms an interrupt occurs which is served for 100 μs.
1ms = 1000μs
Thus, for every 1000μs, (1000 - 100) = 900 μs of main program and 100μs of interrupt overhead exists.
Thus, 900/1000 is usage of CPU to execute main program .
∴ % of CPU to execute main program is (900/1000) × 100 = 90%

0, because global m is not modified, m is just passed to formal argument of P.

0, because n is just passed to formal parameters of Q and no modification in global n.

Since m is defined global it is visible inside all the procedures.

Exchanges a and b.
Let us consider a=5; b=2
a := a+b = 5+2 = 7
b := a-b = 7-2 = 5
a := a-b = 7-5 = 2
O/P: a=2; b=5

Note: Out of syllabus.

Pass 1:
1) Assign address to all statements in the program.
2) Save the values assigned to all tables for use in pass 2.
3) Perform some processing of assembler directives.

In a program the loader that can loads the object of the program from secondary memory into the main memory to execute the corresponding program. Then the loader is to be resides in the main memory.

Root directory can points to the various user directories. Then they will be stored in a way that user can't be easily modify them. Then they should be at fixed location on the disk.

A requires 3, B-4, C-6;
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.

A requires 3, B-4, C-6;
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.

Algorithm: Round robin; TQ: 1

p is departure at 11.

If a graph have n vertices and n edges (n>2) then it is to be cyclic graph. Then it have atleast one cycle and if we remove two edges then it is not connected.
For example let us consider, n=3

p ∧(~p ∨ q)
(p ∧ ~p) ∨ (p ∧ q)
F ∨ (p ∧ q)
(p ∧ q)

Given sets are finite union of sets. One set must be infinite to make whole thing to be infinite.

Relation < is:
1) not reflexive
2) irreflexive
3) not symmetric
4) Asymmetric
5) Antisymmetric

Let,

A one-one function 'f' assigns each element a_i of A a distinct element, b_j=f(a_i) of B_i for a, there are n choices, for a₂ there are n-1 choices, for a_m there are (n-(m-1)) choices.
i.e.,

⇒ In this 'n' is constant. So, n is added to n times itself which is O(n²).
Hence, (a) is wrong. And rest (B), (C), (D), (E) are correct.