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This mock test of Linear Algebra NAT Level - 2 for Physics helps you for every Physics entrance exam.
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*Answer can only contain numeric values

QUESTION: 1

If (a, 6) are the values for which the given equations are inconsistent.

x + 2y + 3z = 4

x + 3y + 4z = 5

x + 3y + az = b

then the value of a is :

Solution:

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

*Answer can only contain numeric values

QUESTION: 2

If z = 7c, then find the value of x such that the given equations has infinite solution

x + 3y – 2z = 0

2x – y + 4z = 0

x – 11y + 14z = 0

Solution:

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

–7

Let

If we take ** c = 7**, we get the solution as

The correct answer is: -10

*Answer can only contain numeric values

QUESTION: 3

A matrix M has eigenvalues 1 and 4 with corresponding eigenvectors (1, –1)^{T} and (2, 1)^{T} respectively. Then M is given by Find the value of α

Solution:

As the given eigen vectors of M are linearly independent, M is diagonalisable,

∴ matrices P and D such that,

or D = P^{–1}MP

M = PDP^{–1}

The correct answer is: 2

*Answer can only contain numeric values

QUESTION: 4

The only real value of λ for which the following equations have non-zero solution is,

x + 3y + 3z = λx

3x + y + 2z = λy

2x + 3y + z = λz

Solution:

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,

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

*Answer can only contain numeric values

QUESTION: 5

For what value of c the following given equations will have infinite number of solutions?

–2x + y + z = 1

x – 2y + z = 1

x + y – 2z = c

Solution:

Consider the augmented matrix,

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

The correct answer is: -2

*Answer can only contain numeric values

QUESTION: 6

Find the rank of the matrix for the following equations :

3x + 4y – z – 6w = 0

2x + 3y + 2z – 3w = 0

2x + y – 14z – 9w = 0

x + 3y + 13z + 3w = 0

Solution:

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

∴ The rank of the matrix is 2.

The correct answer is: 2

*Answer can only contain numeric values

QUESTION: 7

Let A be a 3 × 3 matrix. Suppose that the eigenvalues of A are –1, 0, 1 with respective eigenvectors (1,–1,0)^{T}, (1,1,–2)^{T} & (1,1,1)^{T}. Then 6A is given by Find the value of α

Solution:

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

In this case,

∵ D = P^{–1}AP

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

Hence, **6 A** =

The correct answer is: 5

*Answer can only contain numeric values

QUESTION: 8

Find the rank of the matrix for the following equations :

4x + 2y + z + 3u = 0

6x + 3y + 4z + 7u = 0

2x + y + u = 0

Solution:

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

*Answer can only contain numeric values

QUESTION: 9

Solution of the system,

3x – 2y – w = 2

2y + 2z + w = 1

x – 2y – 3z + 2w = 3

y + 2z + w = 1

is given by (1, 0, 0, α)^{T}. Find the value of α.

Solution:

Consider the augmented matrix,

So, the given system reduces to

5*z* – 9*w* = –9 ....(2)

2*y* + 2*z* + *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

*Answer can only contain numeric values

QUESTION: 10

Let a real symmetric matrix. P is the orthogonal matrix such that P^{-1} AP is a diagonal matrix given by Find the value of α

Solution:

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.,

or

A non zero solution is

Now, again if (

i.e., x – 2y = 0

A non-zero solution is u_{2} = (2, 1)

Now, u_{1} and u_{2} will be orthogonal. Normalising u_{1}and u_{2} yield the orthonormal vectors,

such that

The correct answer is: 6

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