Linear Algebra - 1 - Free MCQ Practice Test with solutions, Maths

We are given that 
T(v) = Av
where A = 
We need to define T(v)
Let v = ,  then T(v) = Av = 
Which reflects v about the x₂-axis.

We are given that the linear transformation T(v) = Av rotates the vectors (-1,0) and (0,1), π radians clockwise.
We need to find the matrix A.
We know that if a vector (a, b) is rotated through an angle a under T. Then


Therefore, the matrix of T is
A = 

We are give that the linear transformation T(v) = Av rotates the vectors v₁ = (-1,0) and v₂ = (0,1) clockwise π radians. We need to find the image of v₁ = (-1, 0) and v₂ = (0,1) under T
We know that if a vector (a, b) is rotated through an angle α under linear transformation T, then
Here v₁ = (-1, 0) and a = π
Therefore,v₂ = (0,1)

= (1,0)
and T(v₂) = T(0, 1)

= (0,-1)

We are given that the linear transformation T(v) = Av rotates the vector v₁ = (-1, 0) and v₂ = (0,1) clockwise π/2radians.
We know that if a vector (a, b) is rotated through an angle a clockwise direction under linear transformation T.Then

Here, v₁ = (-1 , 0) and a = π/2 radians
Therefore, T(v₁) = T(-1, 0)
 =  = (0,1)
Therefore, T(-1, 0) = (0,1)
Similarly, for v₂ = (0, 1) and α = π/2,
and T(v₂) = T(0 ,1)

= (1,0)
Therefore, T(0,1) = (1 ,0 ).

We are given that the linear transformation T(v) = Av rotates the vectors (-1, 0) and (0, 1) clockwise π/2 radians. We need to find the matrix A.
We know that if a vector (a, b) is rotated through an angle a clockwise under T. Then

Here, (a, b) = (-1, 0)  and a = π/2
Therefore, T (-1, 0)

= (0, 1) = 0(-1, 0) + 1(0, 1)
and (a, b) = (0,1), α = π/2

= (1, 0) = - 1(-1, 0) + 0(0, 1)
Therefore, the matrix of T is A = 

We need to find the value o f k where the matrix
A = satisfies the equations A² - kA + 2I = 0.
But A = 
So, 
Putting these values in the equation
A² - kA + 2I= 0, we get
or 
or 1 - 3k + 2 = 0 => k = 1

We are given that the system of equations 

Reduce this system of equations to echelon form, using the operations "R₂ ----> + R₂ - 2R₁" and R₃ ---> R₃ - R₁.
These operations yields:

Also we have, this system has a unique solution so a - 8 ≠ 0
or a ≠ 8

We need to find the determinant of the given matrix,

determinant of A = 
or det (A) = 
det (A) = 1/n!

We are given that the system of equations,
x - 2y - 3z = l, (p + 2)z =3, (2p + l) y + z = 2 is inconsistent. Then we need to find the value of p.
The augmented matrix of this system of equations is,

The solution of this matrix is inconsistent if
p + 2 = 0
or p = -2

We are given that x, y, z are in A.P. with common difference d i. e.,
or 
and the rank of is z.
We need to find the values of d and k. The given matrix can be written as,

Applying R₂ -> R₂, R₁ and R₃ --> R₃ -R₂

Applying, R₃ ---> R₃ - R₂,



 

We have

or equivalently
T(x,y) = (x,y)

 We are given that a linear transformation defined by
T(x, y) = (x + 2y, x - 2y)
Let B ={(1, 0), (0, 1)} be a standard basis then
T(1,0) = (1,1) = 1 (1 ,0)+ 1(0,1)
T(0, 1) = (2 ,-2) = 2(1,0) - 2(0 ,1)
Therefore, the matrix of T with respect to the basis B is

We are given that a transformation defined by
T(x,y,z) = (x,y, 0)
Let α and β be scalars and (x₁, y₁, z₁) and (x₂, y₂, z₂) be two vectors. Then 
T[α(x₁, y₁, z₁) + β(x₁, y₂, z₂)]
or equivalently

Hence, T is linear.

is option D : All of R²

W and T2: W --- X are linear transformations.

The composition of T2 with T1, denoted as T2 ∘ T1, is a linear transformation from V to X.

In other words, for any vector v in V, (T2 ∘ T1)(v) = T2(T1(v)) is a vector in X.

To prove that T2 ∘ T1 is a linear transformation, we need to show that it satisfies the two properties of linearity:

1. Additivity: For any vectors u and v in V, (T2 ∘ T1)(u + v) = T2(T1(u + v)) = T2(T1(u) + T1(v)) = T2(T1(u)) + T2(T1(v)) = (T2 ∘ T1)(u) + (T2 ∘ T1)(v).

2. Homogeneity: For any scalar c and vector v in V, (T2 ∘ T1)(cv) = T2(T1(cv)) = T2(cT1(v)) = cT2(T1(v)) = c(T2 ∘ T1)(v).

Therefore, T2 ∘ T1 is a linear transformation from V to X.

We need to find the dim Horn (R³, R⁴).
Since dim R³ = 3 and dim R⁴ = 4
Therefore,dim Horn (R³, R⁴)
= 3 x 4 = 12
Let U and V be n and m dimensional vector space.
Then dim Hom (U, V) = nm

We are given that C is a non-singular matrix and
B = or B is similar to The eigen values of  are all zeroes. 

Hence, B^k = 0
For       k = 3
            B³ = 0

We are given that the matrix
X=
 
We need to find the rank of X^TX. The transpose of X,
X^T=
 
Now, 
           
Rank (X^TX) = 3

We are given that the system of equation

does not have a solution. We need to find the value of k. The given linear system of equation can be written as,

This system of equation has no. solution if rank (A) # rank (aug A) the augmented matrix is,

This augmented matrix has no. solutions if k = -2.

We are given that the system of equation

It may be written as

Here, the rank of the given coefficient matrix ≤ 3 and number of unknowns are four.
Hence, rank < no. of unknowns this system of equations has infinitely many solutions.