Test: Linear Algebra - 6 - Mathematics MCQ

We need to find the value of k for which the following simultaneous equations
x + y + z= 3
x + 2y + 3z = 4
x + 4y + kz = 6
will not have a unique solution. This system of equation may be written as,

Reduce this system of equation to echelon form using the operations
"R₂ → R₂ - R₁" and R₃ → R₃ -  R₁.
These operations yield -

and also, R₃ → R₃ - 3R₂ gives,

also this system of equation has not unique solution if,
k-7 = 0
k = 7

We need to determine the solution for the system of equation
4y+3z= 8
2x - z = 2
3x + 2y= 5
This system of equation may be written in matrix form as,

Reduce this system of equation to echelon form using the operations
"R₂ ⇔ R₁". This operations yields

and also applying "R₃ → 2R₃ - 3R₁" which yields

again, "R₁→ R₃ - R₂ " we get

Here the coefficient matrix 
and augmented matrix is given by 
Since, the rank of coefficient matrix ≠ the rank of augmented matrix.
Therefore, the solution does not exist.

We are given that the system of equation

This system of equation can be written in augmented matrix as,

Applying the operations "R₂ → R₂ - 2R₁" and "R₃ → R₃ - R₁" these operations yields:

also applying R₃ → 3R₃ - R₂, which yields -

Since the rank of coefficient matrix ≠ the rank of augmented matrix.
Therefore, the given system of equations has no solution.

We are given that A be any 2 x 2 matrix which satisfy A² - A = 0
The characteristic equation of the given matrix is

There exists distinct eigen values.
Hence, A must be diagonalizable.

We are given that A is any n x n with entries equal to 1. Suppose, if A is 2 x 2 matrix with entries equal to 1, that is,

multiplicity of 0 is 1.
For 3 x 3 matrix with entries equal to 1,
i.e.,
 
multiplicity of 0 is 2.
By continuing this process, for n x n matrix with entries equal to 1, that is,
A = 
multiplicity of 0 is (n - 1).

We are given that the system of equation,
x + y + z = 0
x - y - z = 0
This system of equations has rank 2 and 3 unknowns,
that is, rank < no. of unknown
Hence, this system of equation has infinitly many solution.

We are given that a linear transformation T : R³→ R³ defined by
T(1, 0) = (2, 3,1)
and T(l, 1) = (3, 0, 2)
We need to find the image of (x, y) under linear transformation T.
Let there exist scalars α and β such that
(x,y) = α(1,0) + β(1, 1) 
or equivalently (x, y)= (α + β, β)
Comparing the components on both sides, we get
x = α + β and y = β
Solving for α and β, we get
α = x - y and β = y
Therefore,(x, y)= (x - y) (1, 0) + y(1, 1) Taking the image under linear transformation T. we get

implies T(x, y) = (2x + y, 3x - 3y, x+y).

We need to find the linear transformation T : R² → R² such that
T(1, 2) = (2,3) and T(0, 1) = (1, 4).
Let (x, y) ∈ R² and α, β are scalars such that
(x,y) = α(1 , 2) + β(0 ,1)
(x,y) = (α , 2α + β)
On comparing the Camponents of coordinates, we get
α = x and y² = 2α + β
Solving for α and β, we get
α = x and β = y - 2x
Therefore,(x, y) = x (1, 2) + (y - 2x) (0, 1)
Taking the image under linear  transformation T, we get
T(x, y) = T[x(1, 2 ) + (y - 2x) (0,1)]
Using the property of linearity, we get
T(x, y) = xT(1, 2) + (y - 2x) T(0, 1)
Substituting the value of T(1, 2) and T(0, 1), we get
T(x, y) = x(2, 3) + (y - 2x) (1, 4)
           = (2x, 3x) + (y - 2x, 4y - 8x)
           = (y, 4y-5x)

We are given a linear transformation T : R¹⁰ → R⁶ and ker T has dimension 5. We need to find the dimension of range T.
using rank nullity theorem, we get dim range T = dim R¹⁰ - dim ker T.
= 10 - 5 = 5
Therefore, the dim of range T is 5.

We are given that a linear transformation T : R³ → R² defined by T(x, y, z) = (x + y, y + z)
Here, we need to check its linearity and also need to find ker T.
Let (x₁, y₁, z₁) and (x₂, y₂, z₂) be any two vector of R³, Then
T[(x₁, y₁, z₁) + (x₂, y₂, z₂)]

Again let α be a scalar number then,
T[α(x,y, z)]
Hence, T is a linear.
Let (x, y, z) ∈ ker T, then T(x, y, z) = (0,0)
using the definition of ker T, we get
(x+y,y + z) = (0, 0)
Implies x + y - 0 and
            y + z = 0
Implies x = - y and y = - z
Therefore, ker T = {(x, -x, x ) : x e R}
Therefore, dim ker T = 1
Hence, ker T is a proper subspace of R³

Let P₂(x) be the vector space of all polynomials over R of degree less than or equal to 2 and D be the differential operator defined on P₂[x].
We need to find the matrix of D related to the basis {x³, 1, x} Now

Therefore, the matrix of D related to the basis {x², 1, x} is 
 

We are given that a linear transformation T: V₂(R) → V₃(R) defined by
T(a, b) = (a + b , a - b, b).
We need to find the nullity of linear transformation T.
Let (a, b) ∈ ker T, Then
T(a,b)= (0,0,0)
Now using the definition of linear transformation, we get
(a + b, a - b , b) = (0, 0, 0)
Comparing the components of the co-ordinates, we get
a + b = 0, a - b = 0 , b = 0
Solving for a, b, we get
a = 0,6 = 0
Therefore,ker T = {(0, 0)}
Thus, dim (ker T) = 0, that is nullity of linear transformation T is zero.

We need to find the transition matrix P from the standard ordered basis to the ordered basis {(1, 1), (-1, 0)}.
We know that two standard ordered basis for R² is {(1, 0), (0, 1)}.
Therefore, Let there exists scalars α and β such that
(1 ,0 )= α (1, 1) + β (-1,0 )
or equivalently (1,0) = (α - β, α )
Comparing the components of the coordinates, we get α - β = 1, α = 0 Solving for α and β, we get α = 0, β = 1 Therefore, (1, 0) = 0(1, 1) - 1(-1, 0) Again let there be scalars a and P such that
(0 ,1) = α(l, 1) + β ( - l , 0) equivalently(0, 1) = (α - β, α)
Comparing the components of the coordinates, we get α - β = 0, α = 1 Solving for α and β we get
α = 1, β = 1.
Therefore, (0, 1) = 1(1,1) + 1(-1, 0)
By using (1,0) = 0(1, 1) - 1(-1, 0)
and (0,1) = 1(1, 1) + 1 (- 1 ,0 )
The transition matrix is 

Let F be any field and let T be a linear operator on F² defined by
T(a, b) = (a + b, a)
We need to find T^-1(a, b).
Let (a, b) ∈ ker T.
Then T(a, b) = (0, 0)
Using the definition of linear transformation, we get
(a + b, a) = (0, 0)
Comparing the components of the coordinates, we get
a + b = 0, a = 0
Solving for α and β, we get a = 0, b = 0.
Therefore,ker T = {(0,0)}
Hence, T is one-one. Since T is a linear transformation on finite dimensional vector space F². Therefore, T is onto. Hence, T is invertible linear transformation.
Let (x,y) be the image of (a,b) under T^-1.
Then (x, y) = T^-1(a, b) or equivalently T(x, y) = (a, b)
Using the definition of linear transformation,
we get (x+y,y)= (a,b)
Comparing the components of the coordinates, we get
x + y = a and y = b
Solving for x and y, we get x= a - b , y = b.
Therefore,T^-1(a, b) = (a - b, b)

We are given that a linear transformation T : R² → R²
such that T(3, l ) = (2,- 4).
and T(1,1) = (0, 2)
We need to find T(7, 8).
Let there exist scalar α and β such that (7, 8)
= α (3,1) + β(1,1)
or equivalently (7,8) = (3α + β, α + β)
On comparing the components of coordinates, we get
3α + β = 7 and α + β = 8
Solving for α and β, we get

Therefore,
Now taking the image under linear transformation T, we get

Substituting the values of T(3, 1) and T(1, 1), we get

               = (-1,2) + (0,17) = (-1,19)
Thus, the image of (7,8) under the linear transformation T is (-1, 19).

We are given that, A be any 3 x 3 matrix which satisfy
A³- A² + A - I = 0
and we need to find A⁴.
A³- A²+ A - I = 0
or A³ = A ² - A + I 
or A⁴ = A³ - A² + A

We need to find the eigen values of the matrix A =  
The characteristics equation of the given matrix A,
|A - λI |= 0
or 
or (4 - λ) (1 - λ) - 4 = 0
or λ² - 5λ = 0
=> λ(λ-5) = 0
=> λ = 0,5
The eigen values of the given matrix is 0 and 5.

We need to find the eigen vector of the matrix

The characteristic equation of the given matrix is,


Put λ = 5 in equation | A - λI | X = 0 that is,

or 
5z =0
-3z +1 = 0
3z - 4f = 0
z = 0, t = 0
and also let x = k₁, y = k₂
Hence, the eigen vector corresponding to the eigen value λ= 5 is_,

Let k₁ = 1, and k₂ = -2, then 

 

We are given that the eigen vectors o f the matrix  are and . We need to find the value o f a + b. The characteristic equation o f the given matrix is, |A - λI| = 0

or (1 - λ)(2 - λ) = 0 or 
λ = 1,2
Put λ = 1 in the equation [ A- λI ] X= 0

or a = 0 and also put X = 2 in the equation [A - λI] X= 0
or 
or 1+2b = 0
or b = 1/2
Now, a + b = 0 + 1/2
a + b = 1/2

We need to find the eigen vector of the matrix A
= 
The characteristic equation of the given matrix is, |A - λI| = 0
or  = 0
or (1 - λ) (2 - λ,) (3 - λ) = 0
λ = 1,2,3
Suppose, X be the eigen vector corresponding to the eigen value λ = 3.
Hence,  (A - λI) X = 0
or 
or -2x + y = 0
-y + 2z = 0
Let x = k
Hence y = 2k and z = k
Now, the eigen vector is,