Linear Algebra NAT Level - 2 - IIT JAM MCQ

Consider the augmented matrix,

For the following to be inconsistent, a – 4 = 0  and  b – 5 ≠ 0
⇒  a = 4  and  b ≠ 5

The correct answer is: 4

The given system of equations is equivalent to the single matrix equation, AX = 0

To get the solution, consider the augmented matrix,

⇒ z  is a free variable. The given system reduces to
x + 3y – 2z = 0
–7x + 8z = 0
Let  z = c,  where c  is an arbitrary number. Then,

If we take c = 7, we get the solution as  (–10  8  7)^T

The correct answer is: -10

As the given eigen vectors of M  are linearly independent, M  is diagonalisable,
∴   matrices P and D  such that,
or  D = P^–1MP
M = PDP^–1

The correct answer is: 2

The given system is,

For the given system of equations to have a non-zero solution, the coefficient matrix, say A, must be of rank less than 3. Thereby, det A must be zero i.e.,


The roots of  are imaginary.
Hence the only real value of λ  for which the system of equations will have a non-zero solution is 6.

The correct answer is: 6

Consider the augmented matrix,


 

The system will have infinite number of solution if  c = –2.

The correct answer is: -2

Let A  be the coefficient matrix, i.e., consider,




∴  The rank of the matrix is 2.

The correct answer is: 2

We know that,    the matrices D and P such that D = P^–1AP is a diagonal matrix.

In this case, 

 

∵  D = P^–1AP

Pre multiplying by P  and post multiplying by P^–1, we get  A = PDP^–1

Hence,   6A = 

The correct answer is: 5

Consider the augmented matrix, [A : O]  where A the coefficient matrix,

interchanging the variables x and z  


∴   The rank of the matrix is 2.

The correct answer is: 2

Consider the augmented matrix,



So, the given system reduces to

5z – 9w = –9          ....(2)
2y + 2z + w = 1     ....(3)
x – 2y – 3z + 2w = 3 ....(4)
from (1) we get, w = 1
from (2) we get, z = 0
from (3) we get, y = 0
from (4) we get, x = 1

The correct answer is: 1

tr(A) = 7 ; |A| = 6
∴ The characteristic equation will be

⇒ λ = 6, 1 are the eigenvalues of  A.
Now, Let (x, y) be the eigenvector corresponding to λ = 6, then,


i.e., –4x – 2y = 0
or  2x + y = 0
A non zero solution is  u₁ = (1, –2)
Now, again if (x, y) is the eigenvector corresponding to λ = 1. Then 

i.e., x – 2y = 0

A non-zero solution is  u₂ = (2, 1)
Now, u₁ and u₂ will be orthogonal. Normalising u₁and u₂ yield the orthonormal vectors,


such that  

The correct answer is: 6