Test: Linear Algebra - 5 - Mathematics MCQ

We are given that the system of equations,
3x + 2y + z = 0 
x + 4y + z = 0 
2x + y + 4z = 0
It can be written in the matrix form as,

Reduce this system to echelon form using the operation
“R₂ -> 3R₂ - R₁ ” and  " R₃ --> 3R₃ - 2R₂".
These operations yield-

and also "R₃ —> 10R₃ + R₂"

which gives, x = 0, y = 0 and z - 0, Hence, th is system of equations has only the trivial solution.

We need to find the rank of the matrix,

Reduce the matrix to echelon form using the operation.
These operations yield.

and also applying R₄ —> 2R₄ - R₂ we have,

So, rank of M = 2.

We are given that the matrix

We need to find all the eigen values of the given matrix.
Since, the characteristic equations is given by,

We need to find the determinant of the matrix,

The determinant of the given matrix,


= (-3) (1) (1 - 4) + (6) (2) (1 - 4)
= 9 - 36 = -27 

Determinant of Matrix A = 0 so it is not invertible.
Since determinant of matrix is zero it eigen value is zero.
After converting the matrix into row echlon form we find that there exist only 1 non zero row hence the rank of the matrix is 1.

We are given that a linear transformation T : R²—> R² is defined by
T(x,y) = (x + y , x - y)

= αT(x_1,y₁) + βT(x₂, y₂)
Hence, T is a linear transformation.
Let (x, y) ∈ ker T Then T(x, y) = (0, 0) Implies(x + y , x —y) = (0, 0)
Implies x +y = 0 and x - y = 0
Solving for x and y, we get
x = 0, y = 0
Hence ker T= {0, 0}
Therefore,
Nullity of T= dim ker T = 0

We are given that a linear transformation
T : R² —> R³ defined by
T(x, y) = (-x -y, 3x + 8y, 9x - 11y)
We need to find the rank and nullity of T.
Let x, y ∈ ker T.
Then
T(x, y) = (0, 0, 0)
Using the definition of T, we get
Comparing the components on both sides, we get
Solving for * and y, we get x = 0, y = 0.
Hence,ker T= {(0, 0)},
Therefore,Nullity of T = dim ker T = 0.
Using Rank Nullity theorem, we get Rank of T = 2 - Nullity of T
= 2 - 0 = 2 .
Hence, Rank and Nullity of T are 2 and 0 respectively.

We are given that a linear transformation T on R² defined by

for all (x₁, x₂) ∈ R².
We need to find the matrix of T^-1 relative to the standard basis for R².
Implies T is one-one.
Since T is a linear transformation form R² into itself, whose dimension is 2. Hence, T is onto. Therefore, T^-1 exist.
Let (x₁,y₁) be the image of (a, b) under T^-1
i.e., T^-1(a, b) = (x₁ x₂)
Implies (a, b) = T(x₁ ,x₂)
Using the definition of T, we get
(a, b) = (x₁ + x₂, x₁ - x₂)
Implies .x₁ + x₂ = a and x₁ - x₂ = b Solving for x₁ x₂, we get

We know that {(1,0), (0,1)} is an standard basis of R².
Then

Therefore, the matrix of T^-1 with respect to standard basis {(1,0), (0,1)} is

We are given that a linear transformation T : R² —> R² defined by
T(1,0) = T(1 ,2)
and T(1,1) = (0, 2)
We need to find the image of (a, b) under linear transformation T.
Let there exist α and β as scalars such that (a, b)= a (1, 0) + P(l, 1) or equivalently(a, b) = (a + β, β)
Implies a = α + β and b = β
Solving for α, β we get α = a - b
and β = b
Therefore,
(a, b) = ( a - b) (1, 0) + b(1, 1)
Taking the image under linear transformation T, we get
T(a, b) = (a - b) T(1, 0) + bT(1, 1) Implies T(a, b) = (a - b) (1, 2) + b(0, 2)
=(a- b, 2a)

We are given that a linear transformation T:V—>V such that
T(A) = AB - BA
Where, V is a vector space of 2 x 2 matrices over the field of Real Number.
and 
We need to find the Nullity of T.
Let  belongs to ker T
Then T(A) = 0
Using the definition of linear transformation T, we get,
AB - BA = 0
Substituting the values of A and B, we get

or equivalently,
or equivalently, 
Implies c=0 and a + b - d = 0
Therefore, the Nullity of T
= Total Number of variables - Number of Restrictions
= 4 - 2 = 2

Correct Answer :- C

Explanation : Det ( A) = 1*-1*2*-2 = 4

Det (B) = 1⁴ - 5² + 5 = 1 (Hence product of its eigenvalues = 1) [take all 1,1,1,1]

Det(A+B) = -1+1 , 1+1, 2+1, -2+1 = 0*2*3*-1 = 0

We need do determine the characteristic polynomial of the 3 x 3 real matrix

which is companion matrix, where
|A| = - c
trace ( A ) = - a
Therefore, the characteristic polynomial is given by,

We are given that the matrix

Let, A² - A - 5I

which is required zero matrix.

We are given that characteristic equations of a matrix M be
λ² - λ - 1 = 0
It can be written in the matrix form as,

We are given that a linear tansformation T : R³---> R³ defined by T(x, y, z) = (x - y , y + 3 z, x + 2y).
We need to find T^-1.
Let (x, y, z) ∈ ker T. Then
T(x,y, z) = (0, 0, 0)
Using the definition of T, we get
(x -y,y + 3z, x + 2y) = (0, 0, 0) Comparing the components on both sides, we get
x - y =0 , y + 3z = 0, x +2y = 0
Solving for x, y and z, we get
x = 0, y = 0 and z = 0
Hence, ker T= {(0, 0, 0)}, implies T is one-one.
Since T : R³ —> R³ is a linear tansformation and one-one. Hence, T is onto.
Therefore, T^-1 exist.
Let (a,b,c) be the image of (x, y, z) under T^-1.
Then
T^-1(x, y, z) = (a, b,c)
Implies (x, y, z) = T(a, b, c)
Using the definition of T, we get
(x, y, z) = (a - b, b + 3c, a +2b)
Comparing the components on both sides, we get
x =a - b, y = b + 3c, z = a + 2b
Solving for a, b and c, we get

Therefore, T^-1(x,y, z)

We are given that M is a 7 x 5 matrix of rank 3 and N is a 5 x 7 matrix of rank 5.
or 
Since,

We are given that the system of linear equation
x + y + z = 3
x - y - z = 4
x - 5y + kz = 6
has an infinite number of solution. Thus we need to find the value of k. The given system of equations may be written as,

Reduce this system of echelon form using the operations R₂ --> R₂ - R₁ and R₃ --> R₃ - R₁. These operations yield.

and also "R₃ ---> R₃ - 3R₂" we have,

Also, we have this system of equation has infinitely many solution, so,
or k + 5 = 0
or k = - 5

We need to determine the value of A will the matrix given below become singular,

If the given matrix is singular
i.e., =0
8(-12) - λ(-24) = 0 
or 24λ, = 96
or λ, = 4

We are given that the matrix 

We need to find the rank of the matrix. Reduce this matrix to echelon form using the operations "R₂ —> 2R₂ - 3R", and "R₃ —> 2R₃ - R₁".
These operations yield

and also, 

Hence, Rank of A = 2