LinearAlgebra
Question 1 
3  
4  
5  
6 
→ Correction in Explanation:
⇒ (1  λ)(2  λ)  2 = 0
⇒ λ^{2}  3λ=0
λ = 0, 3
So maximum is 3.
Question 2 
Assume that multiplying a matrix G_{1} of dimension p×q with another matrix G_{2} of dimension q×r requires pqr scalar multiplications. Computing the product of n matrices G_{1}G_{2}G_{3}…G_{n} can be done by parenthesizing in different ways. Define G_{i}G_{i+1} as an explicitly computed pair for a given parenthesization if they are directly multiplied. For example, in the matrix multiplication chain G_{1}G_{2}G_{3}G_{4}G_{5}G_{6} using parenthesization(G_{1}(G_{2}G_{3}))(G_{4}(G_{5}G_{6})), G_{2}G_{3} and G_{5}G_{6} are the only explicitly computed pairs.
Consider a matrix multiplication chain F_{1}F_{2}F_{3}F_{4}F_{5}, where matrices F_{1}, F_{2}, F_{3}, F_{4} and F_{5} are of dimensions 2×25, 25×3, 3×16, 16×1 and 1×1000, respectively. In the parenthesization of F_{1}F_{2}F_{3}F_{4}F_{5} that minimizes the total number of scalar multiplications, the explicitly computed pairs is/ are
F_{1}F_{2} and F_{3}F_{4} only
 
F_{2}F_{3} only
 
F_{3}F_{4} only  
F_{1}F_{2} and F_{4}F_{5} only

→ Optimal Parenthesization is:
((F_{1}(F_{2}(F_{3} F_{4})))F_{5})
→ But according to the problem statement we are only considering F_{3}, F_{4} explicitly computed pairs.
Question 3 
Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (H), medium (M) and low (L). Let P(H_{G}) denote the probability that Guwahati has high temperature. Similarly, P(M_{G}) and P(L_{G}) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(H_{D}), P(M_{D}) and P(L_{D}) for Delhi.
The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.
Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (H_{G}) then the probability of Delhi also having a high temperature (H_{D}) is 0.40; i.e., P(H_{D} ∣ H_{G}) = 0.40. Similarly, the next two entries are P(M_{D} ∣ H_{G}) = 0.48 and P(L_{D} ∣ H_{G}) = 0.12. Similarly for the other rows.
If it is known that P(H_{G}) = 0.2, P(M_{G}) = 0.5, and P(L_{G}) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______ .
0.60  
0.61  
0.62  
0.63 
The first entry denotes that if Guwahati has high temperature (H_{G} ) then the probability that Delhi also having a high temperature (H_{D} ) is 0.40.
P (H_{D} / H_{G} ) = 0.40
We need to find out the probability that Guwahati has high temperature.
Given that Delhi has high temperature (P(H_{G} / H_{D} )).
P (H_{D} / H_{G} ) = P(H_{G} ∩ H_{D} ) / P(H_{D} )
= 0.2×0.4 / 0.2×0.4+0.5×0.1+0.3×0.01
= 0.60
Question 4 
Let N be the set of natural numbers. Consider the following sets,

P: Set of Rational numbers (positive and negative)
Q: Set of functions from {0, 1} to N
R: Set of functions from N to {0, 1}
S: Set of finite subsets of N
Which of the above sets are countable?
Q and S only  
P and S only  
P and R only  
P, Q and S only

Set of functions from {0,1} to N is countable as it has one to one correspondence to N.
Set of functions from N to {0,1} is uncountable, as it has one to one correspondence to set of real numbers between (0 and 1).
Set of finite subsets of N is countable.
Question 5 
Consider the following statements.

(I) P does not have an inverse
(II) P has a repeated eigenvalue
(III) P cannot be diagonalized
Which one of the following options is correct?
Only I and III are necessarily true
 
Only II is necessarily true  
Only I and II are necessarily true  
Only II and III are necessarily true

Though the multiple of a vector represents same vector, and each eigen vector has distinct eigen value, we can conclude that ‘p’ has repeated eigen value.
If the unique eigen value corresponds to an eigen vector e, but the repeated eigen value corresponds to an entire plane, then the matrix can be diagonalized, using ‘e’ together with any two vectors that lie in plane.
But, if all eigen values are repeated, then the matrix cannot be diagonalized unless it is already diagonal.
So (III) holds correct.
A diagonal matrix can have inverse, So (I) is false.
Then (II) and (III) are necessarily True.
Question 6 
a unique solution at x = J_{n} where J_{n} denotes a ndimensional vector of all 1  
no solution  
infinitely many solutions  
finitely many solutions 
AX = B
As given that
and c_{1}&c_{n} ≠ 0
means c_{0}a_{0} + c_{1}a_{1} + ...c_{n}a_{n} = 0, represents that a_{0}, a_{1}... a_{n} are linearly dependent.
So rank of 'A' = 0, (so if ‘B’ rank is = 0 infinite solution, ‘B’ rank>0 no solution) ⇾(1)
Another condition given here is, 'Σa_{i} = b',
so for c_{1}c_{2}...c_{n} = {1,1,...1} set, it is having value 'b',
so there exists a solution if AX = b →(2)
From (1)&(2), we can conclude that AX = B has infinitely many solutions.
Question 7 
Let u and v be two vectors in R^{2} whose Euclidean norms satisfy u=2v. What is the value of α such that w = u + αv bisects the angle between u and v?
2  
1/2  
1  
1/2 
Let u, v be vectors in R^{2}, increasing at a point, with an angle θ.
A vector bisecting the angle should split θ into θ/2, θ/2
Means ‘w’ should have the same angle with u, v and it should be half of the angle between u and v.
Assume that the angle between u, v be 2θ (thus angle between u,w = θ and v,w = θ)
Cosθ = (u∙w)/(∥u∥ ∥w∥) ⇾(1)
Cosθ = (v∙w)/(∥v∥ ∥w∥) ⇾(2)
(1)/(2) ⇒ 1/1 = ((u∙w)/(∥u∥ ∥w∥))/((v∙w)/(∥v∥ ∥w∥)) ⇒ 1 = ((u∙w)/(∥u∥))/((v∙w)/(∥v∥))
⇒ (u∙w)/(v∙w) = (∥u∥)/(∥v∥) which is given that ∥u∥ = 2 ∥v∥
⇒ (u∙w)/(v∙w) = (2∥v∥)/(∥v∥) = 2 ⇾(3)
Given ∥u∥ = 2∥v∥
u∙v = ∥u∥ ∥v∥Cosθ
=2∙∥v∥^{2} Cosθ
w = u+αv
(u∙w)/(v∙w) = 2
(u∙(u+αv))/(v∙(u+αv)) = 2
(u∙u+α∙u∙v)/(u∙v+α∙v∙v) = 2a∙a = ∥a∥^{2}
4∥v∥^{2}+α∙2∙∥v∥^{2} Cosθ = 2(2∥v∥^{2} Cosθ+α∙∥v∥^{2})
4+2αCosθ = 2(2Cosθ+α)
4+2αCosθ = 4Cosθ+2α ⇒ Cosθ(uv)+2α4 = 0
42α = Cosθ(42α)
(42α)(Cosθ1) = 0
42α = 0
Question 8 
Both (I) and (II)  
(I) only  
(II) only  
Neither (I) nor (II) 
be a real valued, rank = 2 matrix.
a^{2}+b^{2}+c^{2}+d^{2} = 50
Square values are of order 0, 1, 4, 9, 16, 25, 36, …
So consider (0, 0, 5, 5) then Sum of this square = 0+0+25+25=50
To get rank 2, the 2×2 matrix can be
The eigen values are,
AλI = 0 (The characteristic equation)
λ(λ)25 = 0
λ^{2}25 = 0
So, the eigen values are within [5, 5], Statement I is correct.
The eigen values with largest magnitude must be strictly greater than 5: False.
So, only Statement I is correct.
Question 9 
If the characteristic polynomial of a 3 × 3 matrix M over R (the set of real numbers) is λ^{3}  4λ^{2} + aλ + 30, a ∈ ℝ, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.
5  
6  
7  
8 
λ^{3}  4λ^{2} + aλ + 30 = 0 ⇾ (1)
One eigen value is ‘2’, so substitute it
2^{3}  4(2)^{2} + a(2) + 30 = 0
8  16 + 2a + 30 = 0
2a = 22
a = 11
Substitute in (1),
λ^{3}  4λ^{2}  11 + 30 = 0
So, (1) can be written as
(λ  2)(λ^{2}  2λ  15) = 0
(λ  2)(λ^{2}  5λ + 3λ  15) = 0
(λ  2)(λ  3)(λ  5) = 0
λ = 2, 3, 5
Max λ=5
Question 10 
Two eigenvalues of a 3 × 3 real matrix P are (2 + √1) and 3. The determinant of P is __________.
18  
15  
17  
16 
So, For the given 3×3 matrix there would be 3 eigen values.
Given eigen values are : 2+i and 3.
So the third eigen value should be 2i.
As per the theorems, the determinant of the matrix is the product of the eigen values.
So the determinant is (2+i)*(2i)*3 = 15.
Question 11 
Consider the systems, each consisting of m linear equations in n variables.
 I. If m < n, then all such systems have a solution
II. If m > n, then none of these systems has a solution
III. If m = n, then there exists a system which has a solution
Which one of the following is CORRECT?
I, II and III are true  
Only II and III are true  
Only III is true  
None of them is true 
If R(A) ≠ R(AB)
then there will be no solution.
ii) False, because if R(A) = R(AB),
then there will be solution possible.
iii) True, if R(A) = R(AB),
then there exists a solution.
Question 12 
Suppose that the eigenvalues of matrix A are 1, 2, 4. The determinant of (A^{1})^{T} is _________.
0.125  
0.126  
0.127  
0.128 
Given that eigen values are 1, 2, 4.
So, its determinant is 1*2*4 = 8
The determinant of (A^{1})^{T} = 1/ A^{T} = 1/A = 1/8 = 0.125
Question 13 
5  
6  
7  
8 
l_{11} = 2 (1)
l_{11}u_{12} = 2
u_{12} = 2/2
u_{12} = 1  (2)
l_{21} = 4  (3)
l_{21}u_{12}+l_{22} = 9
l_{22} = 9  l_{21}u_{12} = 9  4 × 1 = 5
Question 14 
a=6,b=4  
a=4,b=6
 
a=3,b=5
 
a=5,b=3 
By properties,
⇒ 6=1+a and 7=a4b
⇒ a=5 ⇒ 7=54b
⇒ b=3
Question 15 
6  
7  
8  
9 
⇒ λ^{2}5λ6=0⇒(λ6)(λ+1)=0⇒λ=6,1
∴ Larger eigen value is 6
Question 16 
0  
1  
2  
3 
Method 2: Determinant is unaltered by the operations (i) and (ii)
∴ Determinant of the resultant matrix = Determinant of the given matrix
(Since C_{1},C_{3} are proportional i.e., C_{3}=15C_{1})
Question 17 
{α(4,2,1)  α≠0, α∈R}  
{α(4,2,1)  α≠0, α∈R}  
{α(2,0,1)  α≠0, α∈R}  
{α(2,0,1)  α≠0, α∈R} 
AX = λX ⇒ (A  I)X = 0
⇒ y+2z = 0 and x+2y = 0
⇒ y = 2z and x/(2) = y
∴ x/(2) = y = 2z ⇒ x/(4) = y/2 = z/1 = α(say)
∴ Eigen vectors are {α(4,2,1  α≠0, α∈R}
Question 18 
pq+r = 0 or p = q = r  
p+qr = 0 or p = q = r
 
p+q+r = 0 or p = q = r  
pq+r = 0 or p = q = r 
Question 19 
0  
1  
2  
3 
The finite dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix.
Question 21 
6  
7  
8  
9 
AX = λX
x_{1} + x_{5} = λx_{1}  (1)
x_{1} + x_{5} = λx_{5}  (2)
(1) + (2) ⇒ 2(x_{1} + x_{5}) = λ(x_{1} + x_{5}) ⇒ λ_{1} = 2
x_{2} + x_{3} + x_{4} = λ∙x_{2}  (4)
x_{2} + x_{3} + x_{4} = λ∙x_{3}  (5)
x_{2} + x_{3} + x_{4} = λ∙x_{4}  (6)
(4)+(5)+(6) = 3(x_{2} + x_{3} + x_{4}) = λ(x_{2} + x_{3} + x_{4} ) ⇒ λ_{2} =3
Product = λ_{1} × λ_{2} = 2×3 = 6
Question 22 
If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.  
If the trace of the matrix is positive, all its eigenvalues are positive.  
If the determinant of the matrix is positive, all its eigenvalues are positive.  
If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. 
• The product of the n eigenvalues of A is the same as the determinant of A. •
A: Yes, for sum to be negative there should be atleast one negative number.
B: There can be one small negative number and remaining positive, where sum is positive.
C: Product of two negative numbers is positive. So, there no need of all positive eigen values.
D: There is no need for all eigen values to be positive, as product of two negative numbers is positive.
Question 23 
Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C_{2} → C_{1} + C_{2}
Question 24 
1024 and 1024  
1024√2 and 1024√2  
4√2 and 4√2  
512√2 and 512√2 
The 2×2 matrix =
Cayley Hamilton theorem:
If matrix A has ‘λ’ as eigen value, A^{n} has eigen value as λ^{n}.
Eigen value of
AλI = 0
(1λ)(1+λ)1=0
(1λ^{2} )1=0
1=1λ^{2}
λ^{2}=2
λ=±√2
A^{19} has (√2)^{19}=2^{9}×√2 (or) (√2)^{19}=512√2
=512√2
Question 25 
1, 4, 3  
3, 7, 3  
7, 3, 2
 
1, 2, 3 
Question 26 
x=4, y=10  
x=5, y=8  
x=3, y=9  
x=4, y=10 
Trace = {Sum of diagonal elements of matrix}
Here given that eigen values are 4, 8
Sum = 8 + 4 = 12
Trace = 2 + y
⇒ 2 + y = 12
y = 10
Determinant = 2y  3x
Product of eigen values = 8 × 4 = 32
2y  3x = 32
(y = 10)
20  3x = 32
12 = 3x
x = 4
∴ x = 4, y = 10
Question 27 
0  
either 0 or 1  
one of 0, 1 or 1  
any real number

When a5=0, then rank(A)=rank[AB]<3,
So infinite number of solutions.
But, it is given that the given system has unique solution i.e., rank(A)=rank[AB]=3 will be retain only if a5≠0.
Question 28 
one  
two  
three  
four 
Answer: We have only one matrix with eigen value 1.
Question 29 
S3 and S2  
S1 and S4  
S1 and S3  
S1, S2 and S3 
If determinant of some square matrix is zero then matrix do not have any inverse.
Hence, from the above definitions we can conclude that S1, S2, S3 are true and S4 is false.
Question 30 
5  
7  
2  
1 
(AλI)^{2}I=0 [a^{2}b^{2}=(a+b)(ab)]
(AλI+I)(AλII)=0
(A(λI)I)(A(λ+I)I=0
Let us assume λ1=k & λ +1=k
λ =k+1 λ =k1
⇓ ⇓
for k=5; λ=4 λ =6
k=2; λ=1 λ =3
k=1; λ=2 λ = 0
k=4; λ=5 λ = 3
So; λ=4,1,2,5,6,3,0,3
Check with the option
Option C = 2
Question 31 
{[1,1,0]^{T}, [1,0,1]^{T}} is a basis for the subspace X.  
{[1,1,0]^{T}, [1,0,1]^{T}} is a linearly independent set, but it does not span X and therefore is not a basis of X.
 
X is not a subspace of R^{3}  
None of the above

Question 33 
Determinant of F is zero  
There are an infinite number of solutions to Fx=b  
There is an x≠0 such that Fx=0  
F must have two identical rows 
Fu = Fv
Fu  Fv = 0
F(u  v) = 0
Given u ≠ v
F = 0 (i.e., Singular matrix, so determinant is zero)
Option A is true.
⇾ Fx = b; where F is singular
It can have no solution (or) infinitely many solutions.
Option B is true.
⇾ x ≠ 0 such that Fx = 0 is True because F is singular matrix (“stated by singular matrix rules). Option C is true.
⇾ F can two identical columns and rows.
Option D is false.
Question 34 
a, a, a  
0, a, 2a  
a, 2a, 2a 
Question 35 
0.5  
0.75  
1.5  
2.0 
Question 36 
it will converge  
it will diverse  
it will neither converge nor diverse  
It is not applicable 
1 + 1/2 <= 9
and 3 + 1 <= 10
Question 37 
1 and 1
 
1 and 6  
2 and 5  
4 and 1

A = (2  λ)(5  λ)  (4) = 0
10  7λ+ λ^{2}  4= 0
λ^{2}  7λ + 6 = 0
λ^{2}  6λ  λ + 6 = 0
(λ  6) 1(λ  6) = 0
λ = 1 (or) 6
Question 38 
no solution
 
a unique solution
 
more than one but a finite number of solutions
 
an infinite number of solutions

2(2  20) +1(3 + 5) + 3(12  2)
= 44 + 8 + 30
= 6 ≠ 0
→ A ≠ 0, we have Unique Solution.
Question 39 
1  
0  
1  
2 
determinant = product of diagonal element [upper triangular matrix]
= 1 * 1 * 1 * 1
= 1
Question 40 
power (2,n)
 
power (2,n^{2})
 
power (2, (n^{2} + n)/2)
 
power (2, (n^{2}  n)/2)

A [i] [j] = A [j] [i]
So, we have only two choices, they are either upper triangular elements (or) lower triangular elements.
No. of such elements are
n + (n1) + (n2) + ... + 1
n(n+1)/2
We have two choices, thus we have
2^{(n(n+1)/2)} = 2^{((n2+n)/2) choices i.e., Power (2, (n2+n)/2). }
Question 41 
D^{1}C^{1}A^{1}
 
CDA
 
ADC
 
Does not necessarily exist

ABCD = I
Pre multiply A^{1} on both sides
A^{1}ABCD = A^{1}⋅I
BCD = A^{1}
Pre multiply B^{1} on both sides
B^{1}BCD = B^{1}A^{1}
CD = B^{1}A^{1}
Post multiply A on both sides
CDA = B^{1}A^{1}⋅A
∴ CDA = B^{1}(I)
∴ CDA = B^{1}
Question 42 
infinitely many
 
two distinct solutions
 
unique
 
none

rank = r(A) = r(AB) = 2
rank = total no. of variables
Hence, unique solution.
Question 43 
≤ a+b  
≤ max(a, b)
 
≤ min(Ma, Nb)
 
≤ min(a, b)

→ Such that a row can have maximum of a elements and no row has separate element and for b also same.
→ By combining the both, it should be ≤ (a,b).
Question 44 
x = 6, y = 3, z = 2  
x = 12, y = 3, z = 4  
x = 6, y = 6, z = 4  
x = 12, y = 3, z = 0 
Question 45 
Find its determinant, Determinant = 3.
Now check options, by putting n=1, I am getting following results,
A) 5
B) 7
C) 3
D) 3
(A), (B) can't be the answer.
Now, check for n=2, Determinant = 91 = 8.
Put n=2 in (C), (D)
C) 7
D) 8
So, (D) is the answer.
Question 46 
nH_{n+1} – (n + 1)  
(n + 1)H_{n} – n  
(n + 1)H_{n} – n  
(n+1)H_{n+1} – (n+1) 
Question 48 
0  
1  
2  
infinitely many

This is in the form AX = B
⇒ R(AB) < n [If we want infinitely many solution]
then 1+5α = 0
5α = 1
α = 1/5 There is only one value of α. System can have infinitely many solutions.
Question 49 
4  
2  
1  
0 
Question 50 
0  
n 1  
n^{2}  3n + 2  
Add i^{th} row and j^{th} column if we zero, apply to all row and their corresponding column the total becomes zero.
Question 51 
4  
0  
15  
20 
Question 52 
has unique solution  
has no solutions  
has finite number of solutions  
has infinite number of solutions

Question 55 
11  
48  
0  
24 
Question 56 
Let a = (a_{ij}) be an nrowed square matrix and I_{12} be the matrix obtained by interchanging the first and second rows of the nrowed Identify matrix. Then AI_{12} is such that its first
row is the same as its second row
 
row is the same as the second row of A  
column is the same as the second column of A  
row is all zero 
So, we can see that column 1 and 2 got interchanged.
Question 57 
Let Ax = b be a system of linear equations where A is an m × n matrix and b is a m × 1 column vector and X is a n × 1 column vector of unknows. Which of the following is false?
The system has a solution if and only if, both A and the augmented matrix [A b] have the same rank.
 
If m < n and b is the zero vector, then the system has infinitely many solutions.  
If m = n and b is nonzero vector, then the system has a unique solution.  
The system will have only a trivial solution when m = n, b is the zero vector and rank (A) = n. 
→ Solution can be depends on rank of matrix A and matrix [A B].
→ If rank[A] = rank[A B] then it can have solution otherwise no solution.
Question 58 
if a = b or θ = nπ, is an integer  
always  
never  
if a cos θ ≠ b sin θ

Question 59 
No solution  
Exactly one solution  
Exactly two solutions  
An infinite number of solutions

x & cos(x) are intersecting at only one point.
Question 60 
1  
2  
n  
Depends on the value of a 
Question 61 
A is closed under* but < A, *> is not a semigroup  
Question 62 
AA′ = 1  
A = A^{1}  
AB = BA  
(AB)' = BA 
Question 64 
In a compact single dimensional array representation for lower triangular matrices (i.e all the elements above the diagonal are zero) of size n × n, nonzero elements (i.e elements of the lower triangle) of each row are stored one after another, starting from the first row, the index of the (i, j)^{th} element of the lower triangular matrix in this new representation is:
i + j  
i + j  1  
If we assume array index starting from 1 then, i^{th} row contains i number of nonzero elements. Before i^{th} row there are (i1) rows, (1 to i1) and in total these rows has 1+2+3......+(i1) = i(i1)/2 elements.
Now at i^{th} row, the j^{th} element will be at j position.
So the index of (i, j)^{th} element of lower triangular matrix in this new representation is
j = i(i1)/2
Question 65 
λ^{3} + 2λ^{2}  2 = 0
Using CayleyHamiltonian theorem
A^{3} + 2A^{2}  2I = 0
So, A^{1} = 1/2 (2A  A^{2})
Solving we get,
Question 66 
(0, 0,α )  
(α,0,0)  
(0,0,1)  
(0, ,0 α )  
Both B and D 
So the question as has
(A  λI)X = 0
AX = 0
What x_{1}, x_{2}, x_{3} are suitable?
Which means:
x_{1} times column 1 + x_{2} times column 2 + x_{3} times column 3 = zero vector
Since α is not equal to zero, so x_{3} must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.
Question 67 
A^{4} = I 
(1λ) (1λ) (iλ) (iλ)
= (λ^{2}1) (λ^{2}+1)
= λ^{4}1
Characteristic equation is λ^{4}1 = 0.
According to CayleyHamilton theorem, every matrix satisfies its own characteristic equation, so
A^{4} = I
Question 68 
S1 and S2 are both true  
S1 is true, S2 is false  
S1 is false, S2 is true  
S1 and S2 are both false 
Question 69 
Fill in the blanks. 
Question 70 
The rows are linearly independent  
The columns are linearly independent  
The row are linearly dependent  
None of the above 
Hence, C is the correct option.