Linear Algebra MCQ Level - 2 - IIT JAM MCQ

Then, consider the augmented matrix,

 

∵  Rank A ≠ Rank [A : B]
∴  The system is inconsistent.
The correct answer is: Inconsistent

tr(A) = 0, |A| = 1
∴ The characteristic polynomial will be,
t(t) = t² + 1 = 0

Now, A is a matrix over the field .Then t(t) has no real roots. Thus A has no eigenvalues and no eigen vectors.

So, A is not diagonalisable over . 
Now, if A is a matrix over C. Then,

t(t) = t² – 1
has two roots i and –i

Thus, A  has two distinct eigenvalues i and –i.

Hence A has two independent eigenvectors. Accordingly, A is diagonalisable over  C.

The correct answer is: A is diagonalisable over C but not 

For the unique solution. |A| should be non-zero where A  is the coefficient matrix. i.e.,


 

⇒ 2k² + 8 ≠ 0
⇒ k² ≠ 4
⇒  k ≠ ± 2

The correct answer is: k ≠ ±2

Consider the augmented matrix,

We can see that the pivot elements exists in columns 1, 3 and 4

So, x₂ and x₅  are free variables.

Given, 2x₁ – 3x₂ – 6x₃ – 5x₄ + 2x₅ = 7     ...(1)
x₃ + 3x₄ – 7x₅ = 6     ...(2)
x₄ – 2x₅ = 1       ...(3)
from (3),
x₄ = 1 + 2x₅
from (2),
x₃ + 3(1 + 2x₅) – 7x₅ = 6
x₃ = 3 + x₅

from (1),
2x₁ – 3x₂ –6(3 + x₅) –5(1 + 2x₅) + 2x₅ = 7
⇒  2x₁ – 3x₂ – 18 – 6x₅ – 5 - 10x₅ + 2x₅ = 7
⇒ 2x₁ – 3x₂ – 23 – 14x₅ = 7
⇒ 2x₁ = 30 + 3x₂ + 14x₅

∵  x₂  and x₅ are free variables, Let  x₂ = x₅ = 0
then  (15, 0, 3, 1, 0)^T  becomes one of the solutions.
The correct answer is: (15, 0, 3, 1, 0)^T

Consider the coefficient matrix,

Rank of the coefficient matrix is 3 which is equal to the number of unknowns.

The given system of equations possesses non zero solution.

The correct answer is: Unique solution

Consider the augmented matrix,



∴  Rank of coefficient matrix = Rank of augmented matrix =  Number of unknowns
Hence,  unique solution, the system reduces to,
–3z = – 9  ⇒  z = 3
2y – z = 1  ⇒  y = 2
x + y + z = 6  ⇒  x = 1
∴  The solution is  (1, 2, 3)^T.

The correct answer is: The system is consistent having unique solution,  (1, 2, 3)^T

For the scalars,  a_i1, a_i2, ..........., a_in  (i,= 1, 2,.....,n)
not all zero such that  a_i1x₁ + a_i2x₂ + ...... + a_in x_n = 0
⇔ that  {x₁, x₂,..........,x_n}  is linearly dependent.
⇒ Rank of coefficient matrix will be less than. Hence, minor of order n in the matrix or determinant of the coefficient matrix will be zero.

The correct answer is: |a_ij|_{n × n} = 0

Coefficient matrix to the given system of equation is,

Hence, the rank of the coefficient matrix is 3 which is equal to the number of unknown variables in the system.
∴  The system has a unique solution.

The correct answer is: The given system is consistent and has a unique solution.

Since, M  is upper triangular, M² will also be upper triangular and hence, both of them will be diagonalisable.

The correct answer is: Both M and M²  are diagonalisable