Matrix Representation of Linear Transformations MCQ Quiz in বাংলা - Objective Question with Answer for Matrix Representation of Linear Transformations - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

let T = $\left[\begin{array}{rrr}1& 3& 1\\ 2& 5& -4\\ 1& -2& 2\end{array}\right]$

B = {[1, 1, 0]T, [0, 1, 1]T, [1, 2, 2]T}

Let [1, 2, 1]T = a[1, 1, 0]T + b[0, 1, 1]T + c[1, 2, 2]T  

⇒ a + c = 1; a + b + 2c = 2; b + 2c = 1

Putting b + 2c = 1 in 2nd equation we get

a + 1 = 2 ⇒ a = 1

then a + c = 1 ⇒ c = 0 and

b + 2c = 1 ⇒ b = 1

We get a = 1, c = 0, b = 1

So, [1, 2, 1]T = 1[1, 1, 0]T + 1[0, 1, 1]T + 0[1, 2, 2]T  

Let [3, 5, -2]T = a[1, 1, 0]T + b[0, 1, 1]T + c[1, 2, 2]T  

⇒ a + c = 3; a + b + 2c = 5; b + 2c = - 2

Putting b + 2c = - 2 in 2nd equation we get

a - 2 = 5 ⇒ a = 7

then a + c = 1 ⇒ c = - 6 and

b + 2c = 1 ⇒ b = 13

So, [3, 5, -2]T = 7[1, 1, 0]T + 13[0, 1, 1]T - 6[1, 2, 2]T  

Let [1, -4, 2]T = a[1, 1, 0]T + b[0, 1, 1]T + c[1, 2, 2]T  

⇒ a + c = 1; a + b + 2c = - 4; b + 2c = 2

Putting b + 2c = 2 in 2nd equation we get

a + 2 = - 4 ⇒ a = - 6

then a + c = 1 ⇒ c = 7 and

b + 2c = 2 ⇒ b = - 12

So, [3, 5, -2]T = -6[1, 1, 0]T - 12[0, 1, 1]T + 7[1, 2, 2]T  

Hence we get the required matrix as

$\left[\begin{array}{ccc}1& 7& -6\\ 1& 13& -12\\ 0& -6& 7\end{array}\right]$

(4) is correct

Concept:

Idempotent matrix is diagonalizable.

Explanation:

A2 = A so A is idempotent matrix

Hence A is diagonalizable

(1) is true

Given CA (x) = x3(x + 1)7 

ρ(A) = number of non-zero eigenvalues = 7

η(A) = 10 - 7 = 3

so dim(C(A)) = 7 ⇒ dim$(\mathrm{C}(\mathrm{A}){)}^{\perp }$ = 3

hence dim$(\mathrm{N}(\mathrm{A}){)}^{\perp }$ = 7

(2), (3) are false

Concept:

1. A square matrix is said to be triangularizable if it is similar to a triangular matrix.

2. Let A be a square matrix whose characteristic polynomial factors into linear polynomials, then A is similar to a triangular matrix. i.e., there exists an invertible matrix P such that P^-1 AP is triangular. 

Explanation:

X = {C ∈ GL₃ (ℝ): CAC^-1 is triangular}

and A ∈ M₃ (ℝ) is fixed.

Since CAC^-1 is always similar to A, thus CAC^-1 is triangular iff A is triangularizable.

Thus if A is not triangularizable, then X = ϕ.

So (1) is false.

Since the degree of the characteristic polynomial of A, i.e., the degree of Ch_A (x) is 3.

Hence, it has at least one real root. As X = Ø which implies Ch_A (x) has 3 distinct roots in C.

⇒ A is diagonalizable. Hence option (c) is correct and options (2) and (4) are false

Only (3) is correct

Explanation:

(1) Let T is identity operator implies T^k = I, ∀ k ∈ $\mathbb{N}$.

Option  (1) is false

(2) Consider T such that matrix of T is $\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right]$, then T - I has matrix $\left[\begin{array}{ll}0& 1\\ 0& 0\end{array}\right]$, but (T - I)^2-1 is non-zero.

Option  (2) is false

Given that 1 in the only eigen value of T. 

which means (T - I)ⁿ = 0 

if Aⁿ = 0 then take another matrix B then AⁿB = 0 because (Aⁿ is nilpotent)

By using this Concept.

So (T-I)ⁿ = 0 Hence (T-I)n+1 = (T - I)ⁿ(T - I) = 0

So (3) and (4) are true.

Concept:

Cayley-Hamilton theorem: According to the Cayley-Hamilton theorem, every matrix 'A' satisfies its own characteristic equation.

Characteristic equation: If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

Explanation:

The characteristic equation is

λ3 + λ2 + 2λ + 1 = 0

Now, by Cayley Hamilton theorem

A³ + A² + 2A + l = 0 ⇒ l = -A³ - A² - 2A

Multiplying by A^-1 on both sides, we get

A^-1 = -A² - A - 2l = -(A² + A + 2I)

∴ A-1 = -(A2 + A + 2I)

Option (4) is correct

Concept:

Hermitian Matrix: Any square matrix say A is said to be a Hermitian matrix if A = Aθ where Aθ is the transpose of the conjugate matrix of A.

Note: All the diagonal elements of a Hermitian matrix is purely real.

Skew Hermitian Matrix: Any square matrix say A is said to be a skew Hermitian matrix if A = - Aθ where Aθ is the transpose of the conjugate matrix of A.

Note: All the diagonal elements of a skew Hermitian matrix is purely imaginary or zero.

Explanation:

(1): We have,

a̅_ij = -a_ji

On putting j = i, we get

a̅_ii = -a_ii

⇒ a̅_ii + a_ii = 0

⇒ Real part of a_ii = 0

⇒ a_ii is purely imaginary.

Hence, the element on the principal diagonal Skew-Hermitian matrix are purely imaginary.

Option (1) is correct

(2)  A is Hermitian matrix. So A = Aθ ....(i)

Now, (iA)^θ =  i̅ A^θ = - iA (Using (i))

Hence, iA is Skew-Hermitian matrix.

Option (2) is correct

(3) A is any matrix

Now, (A - A^θ)^θ = A^θ - (A^θ)^θ

= A^θ - A (as (Aθ)θ = A)

= - (A - Aθ)

Hence, A - A^θ is Skew-Hermitian matrix.

Option (3) is correct

Explanation:

Since given matrix A = $\left[\begin{array}{ccc}1& -1& 0\\ 1& -4& 3\\ -2& 5& -3\end{array}\right]$ w.r.t

standard basis {ν₁, ν₂, ν₃}, where ν₁ = (1, 0, 0), ν₂ = (0, 1, 0), ν₃ = (0, 0, 1).

So, T: V → V is given by T(x) = Ax.

Now for option (1):

T(ν3) = $\left[\begin{array}{ccc}1& -1& 0\\ 1& -4& 3\\ -2& 5& -3\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 3\\ -3\end{array}\right]$≠ 0

So option (1) is false.

For option (2):

T(v₁ + ν2) = $\left[\begin{array}{ccc}1& -1& 0\\ 1& -4& 3\\ -2& 5& -3\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ -3\\ 3\end{array}\right]$≠ 0

So option (2) is false.

For option (3):

T(v1 + ν2 + ν3) = $\left[\begin{array}{ccc}1& -1& 0\\ 1& -4& 3\\ -2& 5& -3\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

So option (3) is true.

For option (4):

T(v1 + ν3) = $\left[\begin{array}{ccc}1& -1& 0\\ 1& -4& 3\\ -2& 5& -3\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 4\\ -5\end{array}\right]$

and T(v2) = $\left[\begin{array}{ccc}1& -1& 0\\ 1& -4& 3\\ -2& 5& -3\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}-1\\ -4\\ 5\end{array}\right]$

So T(v1 + ν3) ≠ T(v2)

Thus option (4) is false.

Explanation:

Recall: The index of an n × n nilpotent matrix is always less than or equal to n.

Let $\mathrm{A}=\left[\begin{array}{lll}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}{\mathrm{A}}^2=\left[\begin{array}{lll}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

Now, $(I+A{)}^n={\left[\begin{array}{lll}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}^n=\left[\begin{array}{lll}1& 1& n-1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\ne I$

for any n > 0.

opt (1) - False. 

(2) Here, If $A=\left[\begin{array}{lll}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$ then cal (A) ≠ {0}

opt (2) - false. 

(4) ∵ Index (A) ≤ 3

Then A³ = O_n×n

and O_n×n is diagonalizable matrix.

⇒ A³ is diagonalizable

opt(4)True.

(5) Eigen values of a nilpotent matrix only 'O'

⇒ opt(3)−False. 

Explanation:

Let X₀ =$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, then for A = $\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$, we have 

AX₀ - X₀A  = A

⇒ $\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$ $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$- $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$  = $\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$

⇒ $\left[\begin{array}{cc}a+c& b+d\\ -a-c& -b-d\end{array}\right]$ - $\left[\begin{array}{cc}a-b& a-b\\ c-d& c-d\end{array}\right]$  = $\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$

⇒ $\left[\begin{array}{cc}b+c& 2b+d-a\\ d-2c-a& -b-c\end{array}\right]$ = $\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$

Which gives the two equations  c = 1 - b

2b = 1 - d + a

So, det(X₀) = ad - bc   

= ad - b(1 - b)  

= ad - $\frac{1(1+d-a)}{4(1-d+a)}$ 

= $\frac{(d+a{)}^2-1}{4}$  

Now, Option 1): when det(X₀) = 0, $\frac{(d+a{)}^2-1}{4}=0$  ⇒ (d + a)² = 1,

which is possible.

Option 2): when det(X₀) = 2, $\frac{(d+a{)}^2-1}{4}$ = 2 ⇒  (d + a)2 = 9 , 

which is possible.

Option 3): when det(X₀) = 6, $\frac{(d+a{)}^2-1}{4}$ = 6 ⇒  (d + a)2 = 25,

which is possible.

Option 4) when det(X₀) = 10, $\frac{(d+a{)}^2-1}{4}$ = 10 ⇒ (d + a)2 = 41 ,

which is not possible.

Also, when A = $\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$  and $\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$,   X₀ satisfying

AX₀ - X₀A = A. 

Since A = $\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$

AX₀ - X₀A = A

⇒ $\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$ $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ - $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$ = $\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$

⇒ $\left[\begin{array}{cc}-a+c& -b+d\\ a-c& b-d\end{array}\right]$ -  $\left[\begin{array}{cc}-a+b& a-b\\ -c+d& c-d\end{array}\right]$ = $\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$

⇒ $\left[\begin{array}{cc}-(b-c)& d-a\\ -(d-a)& b-c\end{array}\right]$ = $\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$

⇒ b - c =  - 1   & b - c = 1 , which is not possible. 

Similarly, A = $\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ case does not give a solution X_0.

The correct answer is option (4).

Explanation:

(1): Take m = 2, A = $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$, b₂ = $\left[\begin{array}{c}1\\ 0\end{array}\right]$, b3 = $\left[\begin{array}{c}2\\ 0\end{array}\right]$.

Then, AX = b₁ i.e $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$ = $\left[\begin{array}{c}1\\ 0\end{array}\right]$ has a solution $\left[\begin{array}{c}1\\ k\end{array}\right]$ for any k ∈ R

Also, AX = b₂ i.e. $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$= $\left[\begin{array}{c}2\\ 0\end{array}\right]$ has a solution $\left[\begin{array}{c}2\\ l\end{array}\right]$ for any l ∈ R

But,  $\left[\begin{array}{c}2\\ 0\end{array}\right]$ = 2 $\left[\begin{array}{c}1\\ 0\end{array}\right]$ i.e. b₂ = 2b_3, which contradicts Statement II.

Hence, whenever A is singular, Statement II may not be true. 

Option (1) is false

(2): Take m =2, A = $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$, b₁ = $\left[\begin{array}{c}0\\ 1\end{array}\right]$. 

Then, AX = b₁⇒ $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$.$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$ = $\left[\begin{array}{c}0\\ 1\end{array}\right]$ has NO solution.

So, there exists b₁= $\left[\begin{array}{c}0\\ 1\end{array}\right]$ such that  AX = b₁ has no solution for A singular.

Option (2) is correct

(3): Take m = 3, A = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$, b₁ = $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$, b₂ = $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

  and b₃ = $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

Then AX = b₁ ⇒ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$$\left[\begin{array}{ccc}{x}_1& x_2& x_3\\ x_4& x_5& x_6\\ x_7& x_8& x_9\end{array}\right]$ = $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ has NO solution.

Similarly, both AX = b₂ and AX = b₃ have solutions. 

Hence, both Statements I and II can be true simultaneously.

Option (3) is correct

(4):

For m =2, take A = $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$, b1 = $\left[\begin{array}{c}0\\ 1\end{array}\right]$. 

Then, AX = b1⇒ $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$.$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$ = $\left[\begin{array}{c}0\\ 1\end{array}\right]$ has NO solution.

 Also, take b2 = $\left[\begin{array}{c}1\\ 0\end{array}\right]$, b3 = $\left[\begin{array}{c}2\\ 0\end{array}\right]$.

Then, AX = b1 i.e $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$ = $\left[\begin{array}{c}1\\ 0\end{array}\right]$ has a solution $\left[\begin{array}{c}1\\ k\end{array}\right]$ for any k ∈ R

Also, AX = b2 i.e. $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$= $\left[\begin{array}{c}2\\ 0\end{array}\right]$ has a solution $\left[\begin{array}{c}2\\ l\end{array}\right]$ for any l ∈ R

But,  $\left[\begin{array}{c}2\\ 0\end{array}\right]$ = 2 $\left[\begin{array}{c}1\\ 0\end{array}\right]$ i.e. b2 = 2b3

Hence, at least one of Statements I or II has to be false.

Option (4) is correct