Test: Linear Algebra - 3 - Mathematics MCQ

We are given that the system of equation,

It may be written in the matrix form as,

Here, the coefficient of matrix is,

and the augmented matrix is given by,

Reduce the system equation in echelon form using the operations
 R₁" and
R₃ -> R₃ - R₁
These operation yield:

and also, R₃ --> R₃ - 2R₂

Here, Rank of A = Rank of aug(A) < number of unknowns.
Hence, the given system of equation has infinite many solutions.

The zero vector space, which contains only the zero vector, has a dimension of 0. This is because the dimension of a vector space is defined as the number of vectors in its basis, and the basis for the zero vector space is the empty set. There are no vectors that can form a basis for this space other than the empty set, since any non-zero vector would not satisfy the requirement that all linear combinations of basis vectors must be able to produce all vectors in the space (in this case, only the zero vector).

So, the correct answer is:

C. 0

We need to find the dimension of Horn (M_2,3, R⁴).
Since dim M_2,3 = 2 x 3 = 6
and dim R⁴ = 4
Therefore,dim Horn (M_2,3, R⁴)
= 6 x 4 = 24
Let M_n,m be the vector space of all n x m matrices over R. Then
dim M_n,m = n x m

 Let T : R³--> R² be the linear transformation defined by
T(x, y, z) = (x, y)
we need to determine the matrix of linear transformation T w.r.t. standard basis of R³ and the basis {(1, 0), (1 ,1 } of R². since {(1 ,0 ,0) , (0,1,0) (0,0,1)} be the standard basis of R³, therefore,

Thus, the matrix of linear transformation T w.r.t. the standard basis R³ and the basis

Let T : R³ -> R³ be the linear transformation defined by

and S : R² ---> R² be the linear transformation defined by
S(x, y) = (2x, 3y)
Now,ker T = {(x, y , z) such that T(x,y, z) = 0}
= {(x, y ,z) : (x, y, 0) = (0,0,0)}
= {(0,0,z) : z ∈ R }
therefore, ker T ≠ {0, 0, 0} hence T is singular.
Next,ker S = {(x, y ) : S(x, y) = 0}
= {(x,y) : (2x,3y) = (0, 0)}
= {(0, 0)}
Therefore, S is non-singular.

We are given that the equation AX = B, where


Hence, the rank of A = 2 and, the augmented matrix is given by,

The rank of A = rank of aug (A) = 2
= number of unknowns.
Hence, there exist a unique solutions.

We are given that the system of equations,
2x + y = 5
x - 3y = - 1
3x + 4y = k is consistent.
We need to find the value of k. The given system of equation may be written as, 
Since, this system of equation is consistent. Therefore

or 2(-3k + 4) -1(k +3) + 5(4 + 9) = 0
-7k + 70 = 0 
k =10

We are given that the system of equation,
x + y + z =0
y + 2z = 0
αx + z= 0
has more than one solution. We need to find the value of α. The determinant of the coefficient matrix must be zero, i.e.,


 

We are given that the system of equations,
x + 2y + z = 9
2x + y + 3z = 7
This system of equation can be expressed in the form AX = B i.e.

which is required matrix form.

We are given that A be an n x n matrix from the set of real num bers and A³ - 3A² + 4A - 6I = 0 ,
where, I is n x n unit matrix.
Since,
A³ - 3A² + 4A = 6I

We are given that a linear transformation T: R²--> R³
show that
T(1,0) = (2,3,1)
and T(1 , 1) = (3, 0, 2)
We need to determine the image of (x, y) under the linear transformation T.
Let there exist scalars α, β,
such that (x,y) = a(l,0) + β(l, 1) or equivalently
(x,y) =(α + β, β)
Comparing the components of the co-ordinates we get,
α + β = x, β = y
Solving for α and β, we get
α = x-y, β = y
therefore, (x, y) = (x - y) (1,0) + y( 1,1) taking the image under linear transformation T, we get

Using the linearity condition, we get T(x,y) = (x-y)T(l,0)+yT(l,l)
Substituting the value of T(1,0) and T(1, 1) we get
T(x, y) = (x -y) (2, 3, 1) + y(3, 0, 2)
= (2x -2y + 3y, 3x - 3y, x - y + 2y)
= (2x + y, 3x - 3y, x + y)
Therefore , the image of (x, y) under T that is

We are given that linear transformation

We need to determine Rank and Nullity of T.
Let (x,y, z, u) ∈ ker T
Then T(x, y, z, u) = (0, 0, 0, 0)
Using the definition of linear transformation we get
(x,y, 0, 0) = (0, 0, 0, 0)
implies x = 0, y = 0, z and u are arbitrary
Therefore,

Hence, Nullity of T = 2
Using Rank Nullity theorem, we get

We are given that the linear transformation T : R³ ---> R³ defined by
We need to determine the rank of the linear transformation T.
Let (x, y, z) ∈ ker T
ThenT(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get,

implies y = 0, z = 0 and x is arbitrary Therefore, ker T = {( x, y, z ) : y = 0, z = 0 and x is arbitrary}
Hence, Nullify of T= 1
Using Rank Nullity theorem, we get Rank T= dim R³ - Nullity T
= 3 - 1 = 2

Let V be the vector space of all 2 x 2 matrices over the field R of real numbers and matrix

Let T : V —> V be a linear transformation defined by T(A) = AB - BA
We need to determine the dim of ker T.
Let 
Then, take T(A) = 0 implies AB - BA = 0
Substituting the values of matrices A and B,
we get

or equivalently

= 
or equivalently 
implies c =0 and a + b - d = 0 Therefore,

Hence, dim (ker T) = Total number of variables - Number of restrictions = 4 - 2 = 2

We are given that the linear transformation T : R³---> R³ defined by

We need to determine Rank of T.
Now, 
Therefore, dim of ker T = 1
Hence by Rank Nullity theorem, we get Rank T = dim R³ - Nullity T
             = 3 - 1 = 2

We are given that a linear transformation T : R³ —> R³ defined by

Therefore
(0,0,1) ∈ ker T and Hence, the Null space is generated by (0, 0,1).

We are given that a linear transformation T : C -->C defined by T(z) =
Where C is a vector space over the field of Real Number R and z is the conjugate of complex number z.
ker T= {z : T(z) = 0}
        = { z : z = 0} = {0}
Hence, T is one-one.
Since C over R is a finite dimensional vector space (two dimensional vector space) and T : C --> C is one linear transformation. Therefore T is onto.

let us consider a 2 × 3 matrix

Where R1 ≠ R2 rank is 2
Another 2 × 3 matrix

Here, R1 = R2 rank is 1
And the rank of these two matrices is 1, 2
So rank cannot be more than m.

 I. We are given that the transformation T: R³ —> R² defined by
T (x , y , z) = (x + 1, y + z)
We need to determine the linearity of the given linear transformation.
Now,T'(0, 0, 0) = (1, 0)
Since the image of (0, 0,0) under transformation T is (1, 0) which is not the zero of R².
Hence, T is non-linear.
II. We are given that the transformation T : R³ --> R defined by
T(x,y,z) = (xy)
We need to determine the linearity of this given transformation.
Since the image of (x, y, z) under linear transformation.
T is an algebraic term of degree 2.
Therefore, T is non linear.
III. We are given that the transformation T: R³ —> R² defined by
T(x, y, z) = (| x |, 0)
We need to determine the linearity of this linear transformation
Let (1, 0, 0) and (-1, 0, 0) be two vectors of R³
Then T(l,0,0) = (l,0)
andT(-l, 0, 0) = (1, 0)
Now,T [(1, 0, 0) + (-1, 0, 0)] =
T(0, 0, 0) = (0, 0)
and T(1, 0, 0) + T(-1, 0, 0)
= (1, 0) + (1,0) = (2, 0)
Hence, T [(1, 0, 0) + (-1, 0, 0)] ≠ T{ 1, 0, 0) + T (-1, 0, 0).
Therefore, T is non linear.

Let T : R² —> R² be a linear transformation such that
We need to determine the image of under linear tansformation T.
Let there exist scalars α and β such that

Since

and taking the image under T, we get