## Linear-Algebra

 Question 1
 A 3 B 4 C 5 D 6
Engineering-Mathematics       Linear-Algebra       Gate 2018
Question 1 Explanation: Correction in Explanation: ⇒ (1 - λ)(2 - λ) - 2 = 0
⇒ λ2 - 3λ=0
λ = 0, 3
So maximum is 3.
 Question 2

Assume that multiplying a matrix G1 of dimension p×q with another matrix G2 of dimension q×r requires pqr scalar multiplications. Computing the product of n matrices G1G2G3…Gn can be done by parenthesizing in different ways. Define GiGi+1 as an explicitly computed pair for a given parenthesization if they are directly multiplied. For example, in the matrix multiplication chain G1G2G3G4G5G6 using parenthesization(G1(G2G3))(G4(G5G6)), G2G3 and G5G6 are the only explicitly computed pairs.

Consider a matrix multiplication chain F1F2F3F4F5, where matrices F1, F2, F3, F4 and F5 are of dimensions 2×25, 25×3, 3×16, 16×1 and 1×1000, respectively. In the parenthesization of F1F2F3F4F5 that minimizes the total number of scalar multiplications, the explicitly computed pairs is/ are

 A F1F2 and F3F4 only B F2F3 only C F3F4 only D F1F2 and F4F5 only
Engineering-Mathematics       Linear-Algebra       Gate 2018
Question 2 Explanation:
→ As per above information, the total number of scalar multiplications are 2173.
→ Optimal Parenthesization is:
((F1(F2(F3 F4)))F5)
→ But according to the problem statement we are only considering F3, F4 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(HG) denote the probability that Guwahati has high temperature. Similarly, P(MG) and P(LG) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(HD), P(MD) and P(LD) 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 (HG) then the probability of Delhi also having a high temperature (HD) is 0.40; i.e., P(HD ∣ HG) = 0.40. Similarly, the next two entries are P(MD ∣ HG) = 0.48 and P(LD ∣ HG) = 0.12. Similarly for the other rows.

If it is known that P(HG) = 0.2, P(MG) = 0.5, and P(LG) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______ .

 A 0.6 B 0.61 C 0.62 D 0.63
Engineering-Mathematics       Linear-Algebra       Gate 2018
Question 3 Explanation: The first entry denotes that if Guwahati has high temperature (HG ) then the probability that Delhi also having a high temperature (HD ) is 0.40.
P (HD / HG ) = 0.40
We need to find out the probability that Guwahati has high temperature.
Given that Delhi has high temperature (P(HG / HD )). P (HD / HG ) = P(HG ∩ HD ) / P(HD )
= 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?

 A Q and S only B P and S only C P and R only D P, Q and S only
Engineering-Mathematics       Linear-Algebra       Gate 2018
Question 4 Explanation:
Set of rational numbers are countable. It is proved by various methods in literature.
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?

 A Only I and III are necessarily true B Only II is necessarily true C Only I and II are necessarily true D Only II and III are necessarily true
Engineering-Mathematics       Linear-Algebra       Gate 2018
Question 5 Explanation: 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 a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1 B no solution C infinitely many solutions D finitely many solutions
Engineering-Mathematics       Linear-Algebra       Gate 2017 set-01
Question 6 Explanation:
ai is a column vector
AX = B As given that and c1&cn ≠ 0
means c0a0 + c1a1 + ...cnan = 0, represents that a0, a1... an 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, 'Σai = b',
so for c1c2...cn = {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 R2 whose Euclidean norms satisfy ||u||=2||v||. What is the value of α such that w = u + αv bisects the angle between u and v?

 A 2 B 1/2 C 1 D -1/2
Engineering-Mathematics       Linear-Algebra       Gate 2017 set-01
Question 7 Explanation: Let u, v be vectors in R2, 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(u-v)+2α-4 = 0
4-2α = Cosθ(4-2α)
(4-2α)(Cosθ-1) = 0
4-2α = 0 Question 8
 A Both (I) and (II) B (I) only C (II) only D Neither (I) nor (II)
Engineering-Mathematics       Linear-Algebra       Gate 2017 set-01
Question 8 Explanation:
Let be a real valued, rank = 2 matrix. a2+b2+c2+d2 = 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,
|A-λI| = 0 (The characteristic equation) -λ(-λ)-25 = 0
λ2-25 = 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 λ3 - 4λ2 + aλ + 30, a ∈ ℝ, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.

 A 5 B 6 C 7 D 8
Engineering-Mathematics       Linear-Algebra       GATE 2017(set-02)
Question 9 Explanation:
For a 3 × 3 matrix ‘M’, the characteristic equation |A – λI| is
λ3 - 4λ2 + aλ + 30 = 0 ⇾ (1)
One eigen value is ‘2’, so substitute it
23 - 4(2)2 + a(2) + 30 = 0
8 - 16 + 2a + 30 = 0
2a = -22
a = -11
Substitute in (1),
λ3 - 4λ2 - 11 + 30 = 0 So, (1) can be written as
(λ - 2)(λ2 - 2λ - 15) = 0
(λ - 2)(λ2 - 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 __________.

 A 18 B 15 C 17 D 16
Engineering-Mathematics       Linear-Algebra       2016 set-01
Question 10 Explanation:
If an eigen value of a matrix is a complex number, then there will be other eigen value, which is conjugate of the complex eigen value.
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 2-i.
As per the theorems, the determinant of the matrix is the product of the eigen values.
So the determinant is (2+i)*(2-i)*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?

 A I, II and III are true B Only II and III are true C Only III is true D None of them is true
Engineering-Mathematics       Linear-Algebra       GATE 2016 set-2
Question 11 Explanation:
i) If m In AX = B,
If R(A) ≠ R(A|B)
then there will be no solution.
ii) False, because if R(A) = R(A|B),
then there will be solution possible.
iii) True, if R(A) = R(A|B),
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 _________.

 A 0.125 B 0.126 C 0.127 D 0.128
Engineering-Mathematics       Linear-Algebra       GATE 2016 set-2
Question 12 Explanation:
Determinant of a matrix is product of the eigen values.
Given that eigen values are 1, 2, 4.
So, its determinant is 1*2*4 = 8
The determinant of (A-1)T = 1/ AT = 1/|A| = 1/8 = 0.125
 Question 13
 A 5 B 6 C 7 D 8
Engineering-Mathematics       Linear-Algebra       GATE 2015 (Set-01)
Question 13 Explanation:
A = LU l11 = 2 -----(1)
l11u12 = 2
u12 = 2/2
u12 = 1 ----- (2)
l21 = 4 ----- (3)
l21u12+l22 = 9
l22 = 9 - l21u12 = 9 - 4 × 1 = 5
 Question 14
 A a=6,b=4 B a=4,b=6 C a=3,b=5 D a=5,b=3
Engineering-Mathematics       Linear-Algebra       GATE 2015 (Set-01)
Question 14 Explanation:
Given λ1=-1 and λ2=7 are eigen values of A
By properties, ⇒ 6=1+a and -7=a-4b
⇒ a=5 ⇒ -7=5-4b
⇒ b=3
 Question 15
 A 6 B 7 C 8 D 9
Engineering-Mathematics       Linear-Algebra       GATE 2015 -(Set-2)
Question 15 Explanation:
Characteristic equation is |4-λ 5 2 1-λ |=0
⇒ λ2-5λ-6=0⇒(λ-6)(λ+1)=0⇒λ=6,-1
∴ Larger eigen value is 6
 Question 16
 A 0 B 1 C 2 D 3
Engineering-Mathematics       Linear-Algebra       GATE 2015 -(Set-2)
Question 16 Explanation: Method 2: Determinant is unaltered by the operations (i) and (ii)
∴ Determinant of the resultant matrix = Determinant of the given matrix (Since C1,C3 are proportional i.e., C3=15C1)
 Question 17
 A {α(4,2,1) | α≠0, α∈R} B {α(-4,2,1) | α≠0, α∈R} C {α(2,0,1) | α≠0, α∈R} D {α(-2,0,1) | α≠0, α∈R}
Engineering-Mathematics       Linear-Algebra       GATE 2015(Set-03)
Question 17 Explanation:
X be an eigen vector corresponding to eigen value λ =1, then
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
 A p-q+r = 0 or p = q = -r B p+q-r = 0 or p = -q = r C p+q+r = 0 or p = q = r D p-q+r = 0 or p = -q = -r
Engineering-Mathematics       Linear-Algebra       GATE 2015(Set-03)
Question 18 Explanation:
For non-trivial solution the value of determinant should be zero. Question 19
The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4-by-4 symmetric positive definite matrix is _____________________.
 A 0 B 1 C 2 D 3
Engineering-Mathematics       Linear-Algebra       GATE 2014(Set-01)
Question 19 Explanation:
For real symmetric matrix, the eigen values are orthogonal to each other. So their dot product will be zero.
The finite dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix.
 Question 20
 A 0 B 1 C 2 D 3
Engineering-Mathematics       Linear-Algebra       Gate 2014 Set -02
Question 20 Explanation: Question 21
 A 6 B 7 C 8 D 9
Engineering-Mathematics       Linear-Algebra       Gate 2014 Set -02
Question 21 Explanation:
Characteristic equation for matrix ‘A’ is
AX = λX x1 + x5 = λx1 ---------- (1)
x1 + x5 = λx5 ---------- (2)
(1) + (2) ⇒ 2(x1 + x5) = λ(x1 + x5) ⇒ λ1 = 2
x2 + x3 + x4 = λ∙x2 -------- (4)
x2 + x3 + x4 = λ∙x3 -------- (5)
x2 + x3 + x4 = λ∙x4 -------- (6)
(4)+(5)+(6) = 3(x2 + x3 + x4) = λ(x2 + x3 + x4 ) ⇒ λ2 =3
Product = λ1 × λ2 = 2×3 = 6
 Question 22
Which one of the following statements is TRUE about every n × n matrix with only real eigenvalues?
 A If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative. B If the trace of the matrix is positive, all its eigenvalues are positive. C If the determinant of the matrix is positive, all its eigenvalues are positive. D If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.
Engineering-Mathematics       Linear-Algebra       Gate 2014 Set -03
Question 22 Explanation:
The sum of the n eigenvalues of A is the same as the trace of A (that is, the sum of the diagonal elements of A).
• 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
 A B C D Engineering-Mathematics       Linear-Algebra       Gate 2013
Question 23 Explanation: Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C2 → C1 + C2   Question 24
Let  be the 2×2 matrix with elements a11 = a12 = a21 = +1 and a22 = -1. Then the eigenvalues of the matrix A19 are
 A 1024 and -1024 B 1024√2 and -1024√2 C 4√2 and -4√2 D 512√2 and -512√2
Engineering-Mathematics       Linear-Algebra       Gate 2012
Question 24 Explanation:
a11=a12=a21=1, a22=-1
The 2×2 matrix = Cayley Hamilton theorem:
If matrix A has ‘λ’ as eigen value, An has eigen value as λn.
Eigen value of |A-λI| = 0 -(1-λ)(1+λ)-1=0
-(1-λ2 )-1=0
-1=1-λ2
λ2=2
λ=±√2
A19 has (√2)19=29×√2 (or) (-√2)19=-512√2
=512√2
 Question 25
 A 1, 4, 3 B 3, 7, 3 C 7, 3, 2 D 1, 2, 3
Engineering-Mathematics       Linear-Algebra       Gate 2011
Question 25 Explanation:
Given matrix is upper triangular matrix and its diagonal elements are its eigen values = 1, 4, 3
 Question 26
 A x=4, y=10 B x=5, y=8 C x=-3, y=9 D x=-4, y=10
Engineering-Mathematics       Linear-Algebra       2010
Question 26 Explanation: 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
 A 0 B either 0 or 1 C one of 0, 1 or -1 D any real number
Engineering-Mathematics       Linear-Algebra       Gate-2008
Question 27 Explanation:
The conjugate matrix [A|B] is When a-5=0, then rank(A)=rank[A|B]<3,
So infinite number of solutions.
But, it is given that the given system has unique solution i.e., rank(A)=rank[A|B]=3 will be retain only if a-5≠0.
 Question 28
 A one B two C three D four
Engineering-Mathematics       Linear-Algebra       Gate-2008
Question 28 Explanation:    Answer: We have only one matrix with eigen value 1.
 Question 29
 A S3 and S2 B S1 and S4 C S1 and S3 D S1, S2 and S3
Engineering-Mathematics       Linear-Algebra       Gate 2008-IT
Question 29 Explanation:
It is given that determinant of square matrix M is zero, means rows of M can be represented as a linear combination of other rows or columns of M can be represented as a linear combination of other columns. Also if determinant of matrix is zero then rank of matrix, which means MX=0 have infinite no.of solution or non – trivial solution.
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
 A -5 B -7 C 2 D 1
Engineering-Mathematics       Linear-Algebra       Gate-2007
Question 30 Explanation:
Eigen value of A[4×4] matrix is -5, -2, 1, 4. |(A-λI)2-I|=0 [a2-b2=(a+b)(a-b)]
|(A-λI+I)(A-λI-I)=0
|(A-(λ-I)I)(A-(λ+I)I|=0
Let us assume λ-1=k & λ +1=k
λ =k+1 λ =k-1
⇓ ⇓
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
 A {[1,-1,0]T, [1,0,-1]T} is a basis for the subspace X. B {[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. C X is not a subspace of R3 D None of the above
Engineering-Mathematics       Linear-Algebra       Gate-2007
Question 31 Explanation:
Since, [1,-1,0]T and [1,0,-1]T are linearly independent set and spans X. So is the basis for subspace X.
 Question 32
 A B C D Engineering-Mathematics       Linear-Algebra       Gate 2007-IT
Question 32 Explanation: Question 33
Fis an n×n real matrix. b is an n×1 real vector. Suppose there are two n×1 vectors, u and v such that, u≠v and Fu=b, Fv=b. Which one of the following statements is false?
 A Determinant of F is zero B There are an infinite number of solutions to Fx=b C There is an x≠0 such that Fx=0 D F must have two identical rows
Engineering-Mathematics       Linear-Algebra       Gate-2006
Question 33 Explanation:
⇾ Fu = b, Fv = b
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 B a, a, a C 0, a, 2a D -a, 2a, 2a
Engineering-Mathematics       Linear-Algebra       Gate 2006-IT
Question 34 Explanation: Question 35
 A 0.5 B 0.75 C 1.5 D 2
Engineering-Mathematics       Linear-Algebra       Gate 2006-IT
Question 35 Explanation: Question 36
 A it will converge B it will diverse C it will neither converge nor diverse D It is not applicable
Engineering-Mathematics       Linear-Algebra       Gate 2006-IT
Question 36 Explanation:
As,
|1| + |1/2| <= |9|
and |3| + |1| <= |10|
 Question 37
 A -1 and 1 B 1 and 6 C 2 and 5 D 4 and -1
Engineering-Mathematics       Linear-Algebra       Gate-2005
Question 37 Explanation: |A| = (2 - λ)(5 - λ) - (4) = 0
10 - 7λ+ λ2 - 4= 0
λ2 - 7λ + 6 = 0
λ2 - 6λ - λ + 6 = 0
(λ - 6) -1(λ - 6) = 0
λ = 1 (or) 6
 Question 38
 A no solution B a unique solution C more than one but a finite number of solutions D an infinite number of solutions
Engineering-Mathematics       Linear-Algebra       Gate-2005
Question 38 Explanation: 2(-2 - 20) +1(3 + 5) + 3(12 - 2)
= -44 + 8 + 30
= -6 ≠ 0
→ |A| ≠ 0, we have Unique Solution.
 Question 39
 A -1 B 0 C 1 D 2
Engineering-Mathematics       Linear-Algebra       Gate 2005-IT
Question 39 Explanation: determinant = product of diagonal element [upper triangular matrix]
= -1 * 1 * 1 * 1
= -1
 Question 40
The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power (2, x) is same as 2x)
 A power (2,n) B power (2,n2) C power (2, (n2 + n)/2) D power (2, (n2 - n)/2)
Engineering-Mathematics       Linear-Algebra       Gate-2004
Question 40 Explanation:
If a matrix is symmetric then
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 + (n-1) + (n-2) + ... + 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
Let A, B, C, D be n × n matrices, each with non-zero determinant. If ABCD = I, then B-1 is
 A D-1C-1A-1 B CDA C ADC D Does not necessarily exist
Engineering-Mathematics       Linear-Algebra       Gate-2004
Question 41 Explanation:
ABCD are n × n matrices with non-zero determinant.
ABCD = I
Pre multiply A-1 on both sides
A-1ABCD = A-1⋅I
BCD = A-1
Pre multiply B-1 on both sides
B-1BCD = B-1A-1
CD = B-1A-1
Post multiply A on both sides
CDA = B-1A-1⋅A
∴ CDA = B-1(I)
∴ CDA = B-1
 Question 42
 A infinitely many B two distinct solutions C unique D none
Engineering-Mathematics       Linear-Algebra       Gate-2004
Question 42 Explanation: rank = r(A) = r(A|B) = 2
rank = total no. of variables
Hence, unique solution.
 Question 43
In an M×N matrix such that all non-zero entries are covered in a rows and b columns. Then the maximum number of non-zero entries, such that no two are on the same row or column, is
 A ≤ a+b B ≤ max(a, b) C ≤ min(M-a, N-b) D ≤ min(a, b)
Engineering-Mathematics       Linear-Algebra       Gate-2004
Question 43 Explanation:
Entry will be a member of same row and same column.
→ 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
 A x = 6, y = 3, z = 2 B x = 12, y = 3, z = -4 C x = 6, y = 6, z = -4 D x = 12, y = -3, z = 0
Engineering-Mathematics       Linear-Algebra       Gate 2004-IT
Question 44 Explanation: Question 45
 A B C D Engineering-Mathematics       Linear-Algebra       Gate 2004-IT
Question 45 Explanation:
Put n=1, you will get a matrix like .
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 = 9-1 = 8.
Put n=2 in (C), (D)
C) 7
D) 8
So, (D) is the answer.
 Question 46
 A nHn+1 – (n + 1) B (n + 1)Hn – n C (n + 1)Hn – n D (n+1)Hn+1 – (n+1)
Engineering-Mathematics       Linear-Algebra       Gate 2004-IT
 Question 47
 A B C D Engineering-Mathematics       Linear-Algebra       Gate 2004-IT
Question 47 Explanation: Question 48
 A 0 B 1 C 2 D infinitely many
Engineering-Mathematics       Linear-Algebra       Gate-2003
Question 48 Explanation: 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
 A 4 B 2 C 1 D 0
Engineering-Mathematics       Linear-Algebra       Gate-2002
Question 49 Explanation:
Number of non-zero rows is the rank of the matrix.
 Question 50
 A 0 B n -1 C n2 - 3n + 2 D Engineering-Mathematics       Linear-Algebra       Gate-2000
Question 50 Explanation:
Let us consider n=5 then we get Add ith row and jth column if we zero, apply to all row and their corresponding column the total becomes zero.
 Question 51
 A 4 B 0 C 15 D 20
Engineering-Mathematics       Linear-Algebra       Gate-2000
Question 51 Explanation:
The value of the determinant is 2 * 1 * 2 * 1 = 4
 Question 52
 A has unique solution B has no solutions C has finite number of solutions D has infinite number of solutions
Engineering-Mathematics       Linear-Algebra       Gate-1998
Question 52 Explanation: Question 53
 A 3 B 1 C 2 D 4
Engineering-Mathematics       Linear-Algebra       Gate-1998
Question 53 Explanation: Question 54
 A a+b B a-b C a+b+c D abc
Engineering-Mathematics       Linear-Algebra       Gate-1998
Question 54 Explanation: Question 55
 A 11 B -48 C 0 D -24
Engineering-Mathematics       Linear-Algebra       Gate-1997
Question 55 Explanation:
Determinant of given matrix = 6×2×4×(-1) = -48
 Question 56

Let a = (aij) be an n-rowed square matrix and I12 be the matrix obtained by interchanging the first and second rows of the n-rowed Identify matrix. Then AI12 is such that its first

 A row is the same as its second row B row is the same as the second row of A C column is the same as the second column of A D row is all zero
Engineering-Mathematics       Linear-Algebra       Gate-1997
Question 56 Explanation:
Let A be 3×3 matrix and I12 be matrix obtained by interchanging the first and second rows of the 3-rowed Identity matrix. 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?

 A The system has a solution if and only if, both A and the augmented matrix [A b] have the same rank. B If m < n and b is the zero vector, then the system has infinitely many solutions. C If m = n and b is non-zero vector, then the system has a unique solution. D The system will have only a trivial solution when m = n, b is the zero vector and rank (A) = n.
Engineering-Mathematics       Linear-Algebra       Gate-1996
Question 57 Explanation:
→ It belongs to linear non-homogeneous equations. So by having m=n, we can't say that it will have unique solution.
→ 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
 A if a = b or θ = nπ, is an integer B always C never D if a cos θ ≠ b sin θ
Engineering-Mathematics       Linear-Algebra       Gate-1996
Question 58 Explanation: Question 59
In the interval [0, π] the equation x = cos x has
 A No solution B Exactly one solution C Exactly two solutions D An infinite number of solutions
Engineering-Mathematics       Linear-Algebra       Gate-1995
Question 59 Explanation: x & cos(x) are intersecting at only one point.
 Question 60
 A 1 B 2 C n D Depends on the value of a
Engineering-Mathematics       Linear-Algebra       Gate-1995
Question 60 Explanation: Question 61
Let A be the set of all non-singular matrices over real number and let* be the matrix multiplication operation. Then
 A A is closed under* but < A, *> is not a semigroup B is a semigroup but not a monoid C is a monoid but not a group D is a group but not an abelian group
Engineering-Mathematics       Linear-Algebra       Gate-1995
Question 61 Explanation:
As the matrices are non-singular so their determinant ±0. Hence, the inverse matrix always exist. But for a group to be Abelian it should follow commutative property. As matrix multiplication is not commutative, is a group but not an abelian group.
 Question 62
Let A and B be real symmetric matrices of size n × n. Then which one of the following is true?
 A AA′ = 1 B A = A-1 C AB = BA D (AB)' = BA
Engineering-Mathematics       Linear-Algebra       Gate-1994
Question 62 Explanation: Question 63
 A 0 B 1 C 2 D 3
Engineering-Mathematics       Linear-Algebra       Gate-1994
Question 63 Explanation: 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, non-zero 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:

 A i + j B i + j - 1 C D Engineering-Mathematics       Linear-Algebra       Gate-1994
Question 64 Explanation:
Though not mentioned in question, from options it is clear that array index starts from 1 and not 0.
If we assume array index starting from 1 then, ith row contains i number of non-zero elements. Before ith row there are (i-1) rows, (1 to i-1) and in total these rows has 1+2+3......+(i-1) = i(i-1)/2 elements.
Now at ith row, the jth 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(i-1)/2
 Question 65
 A B C D Engineering-Mathematics       Linear-Algebra       Gate-1994
Question 65 Explanation:
Using eigen values, the characteristic equation we get is,
3 + 2λ2 - 2 = 0
Using Cayley-Hamiltonian theorem
-A3 + 2A2 - 2I = 0
So, A-1 = 1/2 (2A - A2)
Solving we get, Question 66
 A (0, 0,α ) B (α,0,0) C (0,0,1) D (0, ,0 α ) E Both B and D
Engineering-Mathematics       Linear-Algebra       Gate-1993
Question 66 Explanation:
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.
 Question 67
 A A4 = I
Engineering-Mathematics       Linear-Algebra       Gate-1993
Question 67 Explanation:
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
 Question 68
 A S1 and S2 are both true B S1 is true, S2 is false C S1 is false, S2 is true D S1 and S2 are both false
Engineering-Mathematics       Linear-Algebra       Gate-2001
Question 68 Explanation: Question 69
The number of integer-triples (i.j.k) with 1 ≤ i.j.k ≤ 300 such that i + j + k is divisible by 3 is ________
 A Fill in the blanks.
Engineering-Mathematics       Linear-Algebra       Gate-1991
 Question 70
A square matrix is singular whenever
 A The rows are linearly independent B The columns are linearly independent C The row are linearly dependent D None of the above
Engineering-Mathematics       Linear-Algebra       GATE-1987
Question 70 Explanation:
When the rows are linearly dependent the determinant of the matrix becomes 0 hence in that case it will become singular matrix.
Hence, C is the correct option.
There are 70 questions to complete.