Linear-Algebra
Question 1 |
I. rank(AB) = rank(A) rank(B)
II. det(AB) = det(A) det(B)
III. rank(A + B) ≤ rank(A) + rank(B)
IV. det(A + B) ≤ det(A) + det(B)
Which of the above statements are TRUE?
I and II only
| |
I and IV only
| |
III and IV only
| |
II and III only
|
Rank is the number of independent rows(vectors) of a matrix. On product of two matrices, the combined rank is more than the sum of individual matrices (subrtraced with the order n)
det(AB) = det(A).det(b) as the magnitude remains same for the matrices after multiplication.
Note: We can just take a 2x2 matrix and check the options.
Question 2 |
3 | |
4 | |
5 | |
6 |

→ Correction in Explanation:

⇒ (1 - λ)(2 - λ) - 2 = 0
⇒ λ2 - 3λ=0
λ = 0, 3
So maximum is 3.
Question 3 |
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
F1F2 and F3F4 only
| |
F2F3 only
| |
F3F4 only | |
F1F2 and F4F5 only
|
→ Optimal Parenthesization is:
((F1(F2(F3 F4)))F5)
→ But according to the problem statement we are only considering F3, F4 explicitly computed pairs.
Question 4 |
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 _______ .
0.60 | |
0.61 | |
0.62 | |
0.63 |

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 5 |
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 6 |
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 7 |


a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1 | |
no solution | |
infinitely many solutions | |
finitely many solutions |
AX = B

As given that

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 8 |
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?
2 | |
1/2 | |
1 | |
-1/2 |

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 9 |

(i) One eigenvalue must be in [-5, 5]. (ii) The eigenvalue with the largest magnitude must be strictly greater than 5.Which of the above statements about engenvalues of A is/are necessarily CORRECT?
Both (I) and (II) | |
(I) only | |
(II) only | |
Neither (I) nor (II) |

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 10 |
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 ________.
5 | |
6 | |
7 | |
8 |
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 11 |
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 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 12 |
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(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 13 |
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/ AT = 1/|A| = 1/8 = 0.125
Question 14 |
| 2 2 | | 4 9 |, if the diagonal elements of U are both 1, then the lower diagonal entry l22 of L is
5 | |
6 | |
7 | |
8 |

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 15 |
A = | 1 4 | | b a |
a=6,b=4 | |
a=4,b=6
| |
a=3,b=5
| |
a=5,b=3 |
By properties,

⇒ 6=1+a and -7=a-4b
⇒ a=5 ⇒ -7=5-4b
⇒ b=3
Question 16 |
6 | |
7 | |
8 | |
9 |
⇒ λ2-5λ-6=0⇒(λ-6)(λ+1)=0⇒λ=6,-1
∴ Larger eigen value is 6
Question 17 |

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 C1,C3 are proportional i.e., C3=15C1)
Question 18 |
⎡ 1 -1 2 ⎤ ⎢ 0 1 0 ⎥ ⎣ 1 2 1 ⎦
{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 19 |
px + qy + rz = 0 qx + ry + pz = 0 rx + py + qz = 0then which one of the following options is True?
p-q+r = 0 or p = q = -r | |
p+q-r = 0 or p = -q = r
| |
p+q+r = 0 or p = q = r | |
p-q+r = 0 or p = -q = -r |

Question 20 |
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 22 |
1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 1is ______
6 | |
7 | |
8 | |
9 |
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 23 |
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 24 |
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Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C2 → C1 + C2



Question 25 |
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, 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 26 |

1, 4, 3 | |
3, 7, 3 | |
7, 3, 2
| |
1, 2, 3 |
Question 27 |
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 28 |
0 | |
either 0 or 1 | |
one of 0, 1 or -1 | |
any real number
|

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 29 |
one | |
two | |
three | |
four |




Answer: We have only one matrix with eigen value 1.
Question 30 |
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 31 |
[A I] [I A]where I is the 4 x 4 identity matrix?
-5 | |
-7 | |
2 | |
1 |

|(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 32 |
{[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 R3 | |
None of the above
|
Question 33 |

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Question 34 |
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 35 |
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a, a, a | |
0, a, 2a | |
-a, 2a, 2a |

Question 36 |
0.5 | |
0.75 | |
1.5 | |
2.0 |

Question 37 |
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 38 |
-1 and 1
| |
1 and 6 | |
2 and 5 | |
4 and -1
|

|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 39 |
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 40 |
-1 | |
0 | |
1 | |
2 |

determinant = product of diagonal element [upper triangular matrix]
= -1 * 1 * 1 * 1
= -1
Question 41 |
power (2,n)
| |
power (2,n2)
| |
power (2, (n2 + n)/2)
| |
power (2, (n2 - 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 + (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 42 |
D-1C-1A-1
| |
CDA
| |
ADC
| |
Does not necessarily exist
|
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 43 |
infinitely many
| |
two distinct solutions
| |
unique
| |
none
|

rank = r(A) = r(A|B) = 2
rank = total no. of variables
Hence, unique solution.
Question 44 |
≤ a+b | |
≤ max(a, b)
| |
≤ min(M-a, N-b)
| |
≤ 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 45 |
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 46 |

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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 47 |
nHn+1 – (n + 1) | |
(n + 1)Hn – n | |
(n + 1)Hn – n | |
(n+1)Hn+1 – (n+1) |
Question 49 |
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 50 |
4 | |
2 | |
1 | |
0 |
Question 51 |
0 | |
n -1 | |
n2 - 3n + 2 | |
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Add ith row and jth column if we zero, apply to all row and their corresponding column the total becomes zero.
Question 52 |
4 | |
0 | |
15 | |
20 |
Question 53 |
has unique solution | |
has no solutions | |
has finite number of solutions | |
has infinite number of solutions
|

Question 56 |
11 | |
-48 | |
0 | |
-24 |
Question 57 |
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
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 58 |
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 non-zero 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 59 |
if a = b or θ = nπ, is an integer | |
always | |
never | |
if a cos θ ≠ b sin θ
|

Question 60 |
No solution | |
Exactly one solution | |
Exactly two solutions | |
An infinite number of solutions
|

x & cos(x) are intersecting at only one point.
Question 61 |
1 | |
2 | |
n | |
Depends on the value of a |

Question 62 |
A is closed under* but < A, *> is not a semigroup | |
Question 63 |
AA′ = 1 | |
A = A-1 | |
AB = BA | |
(AB)' = BA |

Question 65 |
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:
i + j | |
i + j - 1 | |
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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 66 |
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-λ3 + 2λ2 - 2 = 0
Using Cayley-Hamiltonian theorem
-A3 + 2A2 - 2I = 0
So, A-1 = 1/2 (2A - A2)
Solving we get,

Question 67 |
(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 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 68 |
A4 = I |
(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 69 |
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 70 |
Fill in the blanks. |
Question 71 |
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.