IIT JAM Mathematics Practice Test- 5 - Mathematics MCQ

On operating [R₁ → R₁-R₂  and R₃→ R₃-R₂]

then

If 

hence rank M = 2

Given

i.e., AA^T = I

A^T = A^-1

we have

Let  and B = [b₁₁ b₁₂ b₁₃..... b_1n] be two non-zero columns and row matrices respectively.

Since A, B are non-zero matrices.

∴ Matrix AB will be a non-zero matrix. The matrix AB will have atleast one non-zero element obtained by multiplying corresponding non-zero elements of A and B. All the two-rowed minors of AB clearly vanish. 

∴ rank of AB = 1

Applying C₁ → C₁ - C₂ + C₃ - C₄, we get

expanding with respect to first column

expanding with respect to the third column

= (a - 1)³ (a - 1)² (a - 1) = (a - 1)⁶.

given

A^-1 = A

⇒ A² = I

Eigen values are given by the solution of equation 

Since x is real and negative , put x = -k, where k is positive constant 

∴ ( 3 - λ ) ( 1 - λ ) + 4 k = 0 

⇒ λ² - 4λ + 3 + 4 k = 0

If λ₂ and λ₂ be the solutions of the above equation, then λ₂ and λ₂  are the eigen values.

Now,Sum of the eigen values = Sum of the roots of the above equation

i.e 

By the well known facts, we know that a Hermitian matrix can have only real eigenvalues and a unitary matrix can have eigenvalues of unit modules. If U be 3 * 3 complex Hermitian matrix which is unitary also then distinct eigenvalues of U are of unit modulus and real. They can be 

Given characteristic equation is λ² - λ - 1 =0, By C - H theorem M² - M - 1 = 0

⇒ M(M - 1) = i

⇒ M^-1 = M - 1

A + I is a triangular matrix. We know that the e igen values of a triangular matrix are its diagonal elements. Eigen values of A² + A + 2 are 3, 7, 13

Row reduce the augmented matrix for (A - 2i)x = 0.

At this point, we are confident that 2 is indeed an eigenvalue of A because the equation (A - 2l)x = 0 have free variables. The general solution is

The eigen space, is

(1,1) Since the matrix is triangular, the eigen values are 1, a, b. If (x₁ x₂, x₃) is an arbitrary eigen vector, (say) corresponding to eigenvalue 1.

We have, x₁ = x_1, ax₂ = X₂ which gives a = 1 and bx₃ = x₃ which gives b = 1. x₂ & x₃ being not zero, 

∴ (a, b) = (1, 1).

Given x + y + z = 3, x + 3y + 2z = 6 and x + λy + 3z = μ

If Δ ≠  0 i.e. λ ≠ 0 the solution is unique.

Case 1 If λ ≠ 5 and μ s any real number, then unique solution exists 

Case 2 If  λ ≠ 5 ⇒ Δ = 0

If Δ₁ ≠ 0 , μ ≠ 9  the system has no solution.
Case 3 If λ = 5 and μ = 9, infinite solutions exist

Interchanging two rows (or columns) change the sign of determinant.

(Operating R₁ -» R₁ + R₃ and R₂ -> R₂ - 2R₃)

When

∴ ρ(A) = 1

where p(A) = number of non-zero rows

when

∴ ρ(A) = 2

when

∴ ρ(A) = 2

, so eigen values are X = 1,1,4

Now  co rresponding to λ = 1 . 

(A - 1) x = 0

Clearly A has only one eigenvector corresponding to λ = 1.

⇒ A is not diagonalizable.

Characteristic eqⁿ will be,  

We know dim V = 3 and S is a linearly independent subset of V which has 3 elements = S form a basis for V

Now take (2x²- 5x + 6) = a(1) + b ( x - 1) + c(x² - 2x + 1) = 3(1) + (-1) ( x -1)+ 2(x² - 2x + 1) = 2x²- 5x + 6 = O ption (C) is correct

we have

Let A be real matrix , then take

(AA^T)^T = (A^T)A^T = AA^T

⇒ AA^T is Symmetric

now

(A^TA)^T = A^T(A^T)^T = A^TA 

⇒ A^TA is Symmetric

Perform ing operations 

we have

Performing Operation wee have

Hence all minors of 3rd and higher orders of the equivalent matrix are zero and one minor  of order 2 whose value = 3 - 4 = -1 ≠0

Hence Rank of the equivalent matrix is Rank (A) = 2.

Given

we know AA^-1 = 1

on Solving (1), (2), (3), (4) we get 

Characterstic eqⁿ of matrix A is

eigen values of matrix A is λ = 3, 9

then eigen values of  A³ → 27 , 729

so  = 27+729 = 756

Hence

or

or

(2 cos x + sin x) (sin x - cos x)² = 0

∴ tan x = - 2 , 1, But  

The characteristic equation of A is 

or (2 - λ) [(6 - λ)( 4 - λ) - 8 ] = 0

or ( 2 - λ) (λ² - 10 λ+ 16) = 0

or ( 2 - λ) (λ - 2 ) ( λ - 8 ) = 0.

There fore the characteristic roots of A are given by λ = 2, 2, 8.
Hence the sum of Distinct Charecteristic Root is 2+8 = 10

since  and if λ is eigen value of A, then λ^-1 is eigen value of A^-1.

Thus for adj (A)X = (A^-1 X) |A| = |A| λ^-1

Thus, eigen value corresponding to λ = 3 is 4/3 and corresponding to λ = -2 is -4/2 = -2

Sum of eigen values = 4/3 - 2

= -2/3

= - 0.67

The system of homogeneous linear equations has a non-trivial solution, if

The augmented matrix is 

Adding the first row to the second row, and 2 times the first row to the third row, we make the entries other than the pivot one in the 1 st column zeros. The row equivalent matrix not looks like

Next, we add 3 times the row to the third row to make the entry below the pivot one in the second column zero. This produces

which is the row echelon form of the augmented matrix. 1st last column is a pivot column unless a = 5. In other words, if the real number a * 5, then the given system is inconsistent.

Here, V has a subspace of dimensions 0, 1 and 2 thus there are three subspaces.

Here, we can se e that vectors (-3 , 0, 1) and (3, 0, - 1 ) are linearly dependent. Hence, dim = 2.