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

Explanation: 

a, b, c, d ∈ ℝ

λx2 + 2xy + y2 = (ax + by)2 + (cx + dy)2 holds for all x, y ∈ ℝ.

⇒ λx2 + 2xy + y2 = a²x² + b²y² + 2abxy + c²x² + d²y² + 2cdxy

⇒ λx2 + 2xy + y2 = (a2 + c2) x2 + 2(ab + cd)xy + (b2 + d2) y2 

Comparing the coefficients of x², y² and xy we get

a2 + c2 = λ ....(i)

ab + cd = 1....(ii)

b2 + d2 = 1 ....(iii)

From (i) λ is the sum of the squares of two integers. So it can't be negative.

Option (1) is false.

If we take a = b = c = d = \(\frac1{\sqrt2}\) then all equations satisfied and

λ = 1

So, options (3) and (4) are false

Hence option (2) is true

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

\(\begin{bmatrix}1&7&-6\\1&13&-12\\0&-6&7\end{bmatrix}\)

(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\(\rm(C(A))^{\perp}\) = 3

hence dim\(\rm (N(A))^{\perp}\) = 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 = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 4}&3\\ { - 2}&5&{ - 3} \end{array}} \right]\) 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) = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 4}&3\\ { - 2}&5&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}} 0\\ 3\\ { - 3} \end{array}} \right]\,\)≠ 0

So option (1) is false.

For option (2):

T(v₁ + ν2) = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 4}&3\\ { - 2}&5&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}} 0\\ -3\\ { 3} \end{array}} \right]\,\)≠ 0

So option (2) is false.

For option (3):

T(v1 + ν2 + ν3) = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 4}&3\\ { - 2}&5&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}} 0\\ 0\\ { 0} \end{array}} \right]\,\)

So option (3) is true.

For option (4):

T(v1 + ν3) = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 4}&3\\ { - 2}&5&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 1 \end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}} 0\\ 4\\ { -5} \end{array}} \right]\,\)

and T(v2) = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ 1&{ - 4}&3\\ { - 2}&5&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 1 \end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}} -1\\ -4\\ { 5} \end{array}} \right]\,\)

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 \(\rm A=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \rm then\ A^2=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \)

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

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 =\(\begin{bmatrix} a&b\\c&d\end{bmatrix}\), then for A = \(\begin{bmatrix} 1&1\\-1&-1\end{bmatrix}\), we have 

AX₀ - X₀A  = A

⇒ \(\begin{bmatrix} 1&1\\-1&-1\end{bmatrix}\) \(\begin{bmatrix} a&b\\c&d\end{bmatrix}\)- \(\begin{bmatrix} a&b\\c&d\end{bmatrix}\)\(\begin{bmatrix} 1&1\\-1&-1\end{bmatrix}\)  = \(\begin{bmatrix} 1&1\\-1&-1\end{bmatrix}\)

⇒ \(\begin{bmatrix} a + c&b+ d\\-a-c&-b-d\end{bmatrix}\) - \(\begin{bmatrix} a - b&a - b\\c-d&c-d\end{bmatrix}\)  = \(\begin{bmatrix} 1&1\\-1&-1\end{bmatrix}\)

⇒ \(\begin{bmatrix} b+ c&2b+ d -a\\d - 2c - a&-b-c\end{bmatrix}\) = \(\begin{bmatrix} 1&1\\-1&-1\end{bmatrix}\)

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 = \(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\)  and \(\begin{bmatrix} 1&-1\\-1&1\end{bmatrix}\),   X₀ satisfying

AX₀ - X₀A = A. 

Since A = \(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\)

AX₀ - X₀A = A

⇒ \(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\) \(\begin{bmatrix} a&b\\c&d\end{bmatrix}\) - \(\begin{bmatrix} a&b\\c&d\end{bmatrix}\)\(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\) = \(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\)

⇒ \(\begin{bmatrix} -a + c& -b + d\\a - c&b-d\end{bmatrix}\) -  \(\begin{bmatrix} -a +b&a- b\\ -c + d&c - d\end{bmatrix}\) = \(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\)

⇒ \(\begin{bmatrix} -(b - c)&d - a\\-(d - a)&b - c\end{bmatrix}\) = \(\begin{bmatrix} -1&1\\1&-1\end{bmatrix}\)

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

Similarly, A = \(\begin{bmatrix} 1&-1\\-1&1\end{bmatrix}\) case does not give a solution X_0.

The correct answer is option (4).