Test: Linear Algebra - 9 - Mathematics MCQ

Let T be the linear operator on R³ defined by
T(x₁ x₂, x₃) = (2x, x₁ - x₂, 5x₁ + 4x₂ + x₃)
Let (x₁, x₂, x₃) ∈ ker T.
Then we need to find T^-1
Now, T(x_1, x₂, x₃) - (0, 0, 0)
Using the definition of T, we get
(2x, x_l - x₂, 5x₁ + 4x₂ + x₃) = (0, 0, 0)
Comparing the components of the coordinates, we get
2x₁ = 0, x₁ - x₂ = 0, 5x₁ + 4 x 2 + x 3 = 0
Solving for x₁ x₂ and x₃, we get x₁ = 0, x₂ = 0, x₃ = 0
Hence,ker T= {(0, 0, 0)}
and hence, T is one-one.
Since R³ is a 3-dimensional vector space and T : R³ —> R³ be a one-one linear transformation and hence, T is onto. Therefore, T^-1 exist.
Let (a, b, c) be the image of (x₁, x₂, x₃) under T^-1.
Then T^-1(x_1,x₂, x₃) = (a, b, c) ImpliesT(a, b, c) = (x_1,x₂, x₃)
Using the definition of T, we get
(2a a - b 5a + 4b + c) = (x_1,x₂, x₃) Comparing the components, we get 2a = x₁, a - b = x₂, 5a l 4b I c = x₃ Solving for a, b, c, we get


Therefore, T^-1(x₁ x₂, x₃) = 

We are given that the matrix
M =
 
We need to find the minimal polynomial of the given matrix. The characteristic polynomial of the given matrix is given by,
Hence, the minimal polynomial of the given matrix M is,

We are given that the eigen values of a 3 x 3 real matrix A are 1,2 and -3 say λ₁ λ₂ and λ₃.
The characteristic equation of the given matrix is,

By Cayley Hamilton’s theorem, we can write this equation in the matrix form as,

We need to find the minimal polynomial of the matrix

The characteristic polynomial of the given matrix is,
ch_A(x) = (x - 1)² (x - 2) (x - 2)
and the minimal polynomial of the given matrix
= l.c.m. of (x - 1)², (x - 2) and (x-2)
= (x-1)² (x-2)

We are given that A be a 3 x 3 matrix whose ' characteristic roots are 3, 2, -1 say λ₁, λ₂ and λ₃ and B = A² - A
λ₁ = 3,
λ₁ = (3)² - 3 = 6
λ² = 2,
λ₂ = (2)² - 2 = 2
λ₃ = -1
λ₃ = (-1)² - (-1) = 2
The determinant of B is, | B | = product of its eigen values
= λ₁'λ₂'λ₃'
= 6 x 2 x 2 = 24
 

We are given that,


and 

We need to find the minimal polynomial of the matrix

The characteristic polynomial of the given matrix M is,
ch_m (x) =(x - 3)² (x - 3) (x + 3)² (x - 0)²
And the minimal polynomial of the given matrix M is
m_x(x) = l.c.m. o f (x - 3)²,
(x - 3),(x + 3)², (x - 0)²
or m_m(x) = (x - 3)² (x + 3)²x²
or m_m (x) = x²(x - 3)² (x + 3)²

We are given that a linear operator T : Cⁿ—> Cⁿ of rank n - 2.
Since T : Cⁿ —> Cⁿ is a linear operator of rank n - 2.
Therefore, T is singular operator, Hence 0 must be an eigen value of T.

Consider the linear transformation defined by
T₁(x,y) = (x,x, 0)
Let (x₁,x₁) and (x₂,x₂​) be two vectors and α, β be scalars. Then
T₁[α(x₁ y₁) + β(x₂, y₂)]
= T₁[(αx₁ + βx₂, αy₁ + βy₂)]
= (αx₁ + βx₂, αx₁ + βx₂, 0)
= α(x₁,x₁, 0) + β(x₂, x₂, 0)
= αT₁(x₁,y₁) + βT₁(x₂,y₂)
or equivalently

Therefore, T₁ is linear transformation.
Let (x,y) ∈ ker T₁.
Then T(x,y) = (0, 0, 0)
Using the definition of linear transformation T₁,we get
(x, x, 0) = (0, 0, 0)
Comparing the components of the co-ordinates, we get,
x = 0 and y is arbitrary
Therefore,ker T₁{ = {(0, y ) : y ∈ R}
Hence,dim ker T₁ = 1.
Using Rank nullity theorem Rank T₁
= 2 - dim ker T₁ 
= 2 - 1 = 1.
Therefore, 'T₁ is a linear transformation of rank 1.
Next, consider the linear transformation T₂ defined by
T₂(x,y) = (x,x+y,y)
Let (x₁,y₁) and (x₂,y₂) be any two vectors and α, β are scalars.
Then Hence, T₂ is a linear transformation.
Let (x, y) ∈ ker T.
Then T₂(x,y) = (0, 0, 0)
Using the definition of linear transformation T₂, we get
(.x, x+ y,y ) = (0, 0, 0)
Comparing the components of the coordinates, we get
x = 0, x + y = 0, y = 0
Solving for x and y, we get x = 0, y = 0
Hence,ker T₂ = {(0,0)}.
Therefore,dim ker T₂ = 0
Using rank nullity theorem,
Rank T₂ = 2 - dim ker T₂
= 2 - 0 - 2 .
Hence, T₂ is a linear transformation of rank 2.
Next, consider the linear transformation T3 defined by
T₃(x,y) = (x,x+1, y)
Then T₃(0, 0) =(0, 1, 0)
but (0,1,0) is not a zero of codomain  space.
Hence, the image of (0, 0) under T3 is not a zero.
Therefore T3 is not a linear transformation.

Let V, W and X be three finite dimensional vector spaces such that dim V = dim X. Suppose S : V --> W and T : W --> X are two linear maps such ToS : V --> X is injective.Then S is injective and T is surjective.

Let S and T be two linear operators on R³ defined by
S(x, y, z) = (x, x + y, x - y - z)
and T(x, y, z) = (x + 2z, y - z, x + y + z).
Let (x, y, z) ∈ ker S.
Then S(x,y,z)= (0,0,0)
Using the definition of linear transformation S, we get
(x , x + y , x - y - z ) = (0, 0, 0) Comparing the components of the coordinates, we get
x = 0 , x+ y = 0, x - y - z = 0
Solving for x, y and z, we get
 x = 0, y = 0, z = 0
Hence,ker S = {(0, 0, 0)}
Therefore, S is one-one.
Since S is a one-one linear operator on R³. Therefore, S is onto and hence invertible that is non-singular.
Next, Let (x, y, z) ∈ ker T. Then
T(x,y,z)=(0,0,0)
Using the definition of linear transformation T, we get
(x + 2z ,y - z , x + y + z) = (0, 0, 0) Comparing the components of the coordinates, we get
x + 2 z = 0 , y - z = 0, x + y + z = 0 Solving for x, y and z, we get x = -2z, y = z Therefore, ker

Hence, dim ker T= 1. Therefore, T is not one-one and hence T is not invertible that is T is singular.

We are given that A be a matrix of quardratic form (x₁-x₂ + 2x₃)²
or x₁² + x₂² + 4x₃² - 2x₁x₂ - 4x₂x₃ + 4x₃x₁
trace (A) = sum of coefficient of
        x₁², x₂² and x₃².
        = 1 + 1 + 4
        = 6

We are given that the eigen values of a 3 x 3 real matrix P are 1, - 2 ,3 say λ₁, λ₂, λ₃.
The characteristic equation of the given matrix P is,
x² - (trace P) x² + (λ₁•λ2 + λ₂•λ₃ + λ₃•λ₁)x - |P| = 0
or x³ - (1 - 2 + 3)x² + (-2 - 6 + 3) x -(-6) = 0
or x³ — 2x² — 5x + 6 = 0
By Cayley Hamilton theorem, we can write this equation in the matrix form.
p³ - 2P² - 5p + 6I = 0 
or 6I = 5P + 2P² - P³
or 6I = P(5I + 2P - P²)
or P^-1 = 1/6(5I + 2P - P²)
 

We are given that the system of equations,
x + y+z= 0
3x + 6y + z = 0
αx + 2y + z = 0
has infinitely many solutions. We need to find the value of a. The given system of equation may be written in matrix form as,

Reduce this system of equation to echelon form using the operations
‘R₂ --> R₂ - R₁' and R₃ ---> R₃ - R₁, these operations yields -

Also applying , R₃ —> 5R₃ - R₂ we get -

Since, this system of equations has infinitely man solution. It is possible when 5α - 7 = 0
or α = 7/5

We are given that the system of equations,
x + y +z = 3
x - y - z = 4
x - 5y + kz = 6
has infinitely many solution and we need to find the value of k. The given system of equation may be written in the matrix form as,

Reduce this system of equations to echelon form using the operations "R₂ —> R₂ - R₁", "R₃ --> R₃ - R₁" these operations yield -

and also applying "R₃ -> R₃ - 3R₂" we get.

Since, the system has infinite solution it is possible when,
k + 5 = 0 or k = - 5