Linear Algebra MCQ Quiz in বাংলা - Objective Question with Answer for Linear Algebra - বিনামূল্যে ডাউনলোড করুন [PDF]

ব্যাখ্যা:

প্রদত্ত:

V = {(a, b, c, d) : b - 5c + 2d = 0}

W = {(a, b, c, d) : a - d = 0, b - 3c = 0}

W থেকে: a - d = 0 ⇒ a = d

W থেকে: b - 3c = 0 ⇒ b = 3c

V থেকে: b - 5c + 2d ⇒ 0

W এর সমীকরণগুলি থেকে a এবং b এর মান V এর সমীকরণে প্রতিস্থাপন করুন:

3c - 5c + 2d = 0

⇒ -2c + 2d = 0

⇒ -2c = -2d

⇒ c = d

এখন আমাদের আছে:

a = d , b = 3c এবং c = d

সুতরাং, V ∩ W এর ভেক্টরগুলিকে এভাবে প্রকাশ করা যেতে পারে:

(d, 3d, d, d) = d (1, 3, 1, 1)

ছেদ V ∩ W একটি একক ভেক্টর দ্বারা বিস্তৃত: (1, 3, 1, 1)

অতএব, এর মাত্রা হল 1

V ∩ W এর মাত্রা হল 1

সুতরাং বিকল্প (1) সঠিক উত্তর।

Concept:

Let V be a vector space. A subset W of V is said to be the subspace of V if α, β ∈ W and a ∈ F ⇒ aα + β ∈ W 

Explanation:

We have, V is the real vector space of all mapping from ℝ R and V1 = {f ∈ V/f(-x) = f(x)}

V2 = {f ∈ V: f(-x) = -f(x)} vector space.

Let V be a vector space over field F and let W \(\subseteq\) V. Then, W is called a subspace of V iff W itself is a vector space over F with respect to operations of vector addition and scalar multiplication in V.

The necessary and sufficient condition for a non-empty subset W of a vector space V(F) to be subspace of V is a, b ∈ F and for all α, β  ∈ W  

⇒ aα + bβ ∈ W --- (1)

To show V₁ and V₂ as subspace of V. Suppose f, g ∈ V₁ and a is any scalar i.e., a ∈ R. Then,

(af + g)(-x) = af(-x) + g( -x)

= a f(x) + g(x)

= (af + g) (x)

⇒ af + g ∈ V₁

Thus, a ∈ R and f, g ∈ V₁ ⇒ af + g ∈ V₁.

Hence, V₁ is a subspace of V.

Again suppose f₀, g₀ ∈ V₂ and a is any scalar, then

(af₀ + g₀)(-x) = af₀(-x) + g₀ ( -x)

= a [ - f₀ (x) -g₀ (x)]

= -[af₀ (x) + g₀(x)]

= - [ af₀ + g₀] x

⇒ af₀ + g₀ ∈ V₂

Hence, V₂ is subspace of V.

Option (4) is true

Concept:

Row operations, also known as elementary row operations, are operations performed on the rows of a matrix that preserve the solution set of a system of linear equations. 

Explanation:

A = \(\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 & 3 \\ 1 & 2 & 3 & 4 & 4 \\ 1 & 2 & 3 & 4 & 5 \end{array}\right)\)

R_i \(\leftrightarrow\) R_i - R₁ {i=2,3,4,5}

A = \(\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 & 2 \\ 0 & 1 & 2 & 3 & 3 \\ 0 & 1 & 2 & 3 & 4 \end{array}\right)\)

Ri \(\leftrightarrow\) Ri - R2 {i=3,4,5}

A = \(\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 2 & 3 \end{array}\right)\)

Ri \(\leftrightarrow\) Ri - R3 {i=4,5}

A = \(\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 2 \end{array}\right)\)

R5 \(\leftrightarrow\) R5 - R4 

A = \(\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right)\)

Thus, ρ(A) = 5

Hence, Option (4) is true 

Explanation:

(-, -) is a symmetric bilinear form on ℝ2 such that there exist nonzero v, w ∈ ℝ2 such that (v, v) > 0 > (w, w) and (v, w) = 0

Let f((x₁, x₂), (y1, y2)) = x1y1 - x2y2

Also, let v = (1, 0) and w = (0, 1)

Then f(v, v) = f((1, 0), (1, 0)) = 1 - 0 = 1 > 0

f(w, w) = f((0, 1), (0, 1)) = 0 - 1 = -1 < 0

f(v, w) = f((1, 0), (0, 1)) = 1 - 1 = 0

So, here A = \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

A² = \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) = \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) ≠ 0

(1) is false

rank(A) = 2

(2), (3) are false

Let u = (1/2, 1/2) ≠ 0  then f((1/2, 1/2), (1/2, 1/2)) = 1/4 - 1/4 = 0

Hence there exists u ∈ ℝ2, u ≠ 0 such that (u, u) = 0. 

(4) is correct

Concept:

A quadratic form is degenerate if at least one eigenvalue is 0 

Explanation:

B = \(\left[\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & -2 \end{array}\right]\) and S = \(\left\{\left[\begin{array}{l} \rm a \\\rm b \\\rm c\end{array}\right] \in \rm ℝ^3 \mid Q(a, b, c)=0\right\}\).

So quadratic form is Q(x, y, z) = x² + y² -2z² + 2xy

Now, Q(a, b, c) = 0

⇒ a2 + b2 -2c2 + 2ab = 0....(i)

(1): On xy-plane, z = 0 so c = 0

Therefore (i) implies

a 2 + b2 + 2ab = 0 ⇒ (a + b)² = 0 ⇒ a + b = 0

i.e., x + y = 0, which is a line

So option (1) is TRUE

(2) On xz-plane, y = 0 so b = 0

Therefore (i) implies

a2 - 2c2 = 0 ⇒ x2 - 2z2 = 0 which is not an equation of ellipse

So option (2) is FALSE

(3): (i) ⇒  a2 + b2 -2c2 + 2ab = 0

  ⇒ (a + b)² = 2c²

  ⇒ a + b = ± √2c

So S is the union of two planes.

Option (3) is TRUE

(4): B = \(\left[\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & -2 \end{array}\right]\) 

eigenvalues of B are -2, 2, 0

So Q is a degenerate quadratic form.

Option (4) is TRUE

Concept:

1) let A = \(\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\)

 then A^T =  \(\begin{bmatrix}a&d&g\\b&e&h\\c&f&i\end{bmatrix}\)​​

Signature of a real quadratic form is express of positive terms over negative term in real quadratic form . 

let p = positive terms

     n = negative terms

then signature s = p - n​

Calculation:

Let A = \(\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\)

then AT =  \(\begin{bmatrix}a&d&g\\b&e&h\\c&f&i\end{bmatrix}\)

given A+AT = K.I

now 

A+AT = \(\begin{bmatrix}2a&b+d&c+g\\d+b&2e&f+h\\g+c&h+f&2i\end{bmatrix}\) . . . . . . . . . . 1

A+AT = K.I

A+AT = \(\begin{bmatrix}K&0&0\\0&K&0\\0&0&K\end{bmatrix}\) . . . . . . . . . . 2

now from equation 1 and 2 , we get 

2a = K, 2e = K and 2i = K

a = e = i = K/2

now put all the values in matrix A ,

A= \(\begin{bmatrix}K/2&b&c\\-b&K/2&f\\-c&-f&K/2\end{bmatrix}\)

Trace A = K/2 + K/2 + K/2

             = \(\frac33\)K

A2 = \(\begin{bmatrix}\frac{k^2}{4} -b^2 -c^2 &X&Y\\Z&\frac{k^2}{4 }-b^2 -f^2&P \\Q&R&-c^2 -f^2 +\frac{k^2}{4}\end{bmatrix}\)

Where 

X = product of first row and second column

Y = product of first row and third column

Z = product of second row and first column

P = product of second row and third column

Q = product of third row and first column

R = product of second row and second column

We don't need these terms so we used variables here .

We need only Trace of A² .

Trace (A2) = \(\frac34\)K2 - 2b2 - 2c2 - 2f2 

given 

q(A) = Trace (A)2 - Trace (A2)

q(A) = (\(\frac32\)K)2 - \(\frac34\)K2 - 2b2 - 2c2 - 2f2 

q(A) = \(\frac32\) K2 + 2b2 + 2c2 + 2f2 

Matrix is 

\(\begin{bmatrix}3/2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&2\end{bmatrix}\)

All eigen values are positive 

signature = ( + , + , + , + )

correct option is 1

Concept:

The generalized coordinate transformation from (p, q) to (P, Q) is canonical if \(\frac{\partial (P, Q)}{\partial (p,q)}\) = 1

Explanation:

Given Q = pq(a+1), P = qb is canonical if \(\frac{\partial (P, Q)}{\partial (p,q)}\) = 1

Now, \(\frac{\partial Q}{\partial p}\) = q(a+1), \(\frac{\partial Q}{\partial q}\) = (a+1)pqa, \(\frac{\partial P}{\partial p}\) = 0, \(\frac{\partial P}{\partial q}\) = bqb-1 

\(\frac{\partial (P, Q)}{\partial (p,q)}\) = 1

⇒ \(\begin{vmatrix}\frac{\partial P}{\partial p}&\frac{\partial P}{\partial q}\\\frac{\partial Q}{\partial p}&\frac{\partial Q}{\partial q}\end{vmatrix}\) = 1

⇒ \(\begin{vmatrix}0&bq^{b-1}\\q^{a+1}&(a+1)pq^a\end{vmatrix}\) = 1

⇒ - bq^a+b = 1

Only option (4) is satisfying the above relation.

Hence option (4) is corret

Solution - We need to check the given matrix not diagonalizable or not 

When all eigen values are distinct then Algebraic Multiplicity is equal to  Geometric Multiplicity which is 1 for option 1) , option 2) and option 4) 

these matrix are upper triangular and lower triangular and the diagonal entries are its eigen values 

So, Option 1) , Option 2) and Option 4) are diagonalizable as eigenvalues are distinct

Now for 

Option 3) \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) 

Here its eigen value be 1,1 and Algebraic Multiplicity be 2 

and Geometric Multiplicity be 1 

So \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) is not diagonalizable.

Therefore, Correct Option is Option 3) 

Alternate Method Every matrix which have distinct eigenvalues are diagonalizable and main diagonal elements are the eigenvalues of upper and lower triangular matrix.

For options (1), (2), (4), each matrix has distinct eigenvalues. So they are diagonalizable.

Hence matrix from option (3) is not diagonalizable

Concept:

A matrix M is positive if v, v ≥ 0, ∀ v ∈ ℝn

Explanation:

α, β v, v > = \(\left\langle\left[\begin{array}{lll} α & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & β \end{array}\right]\left[\begin{array}{l} v_1 \\ v_2 \\ v_3 \end{array}\right],\left[\begin{array}{l} v_1 \\ v_2 \\ v_3 \end{array}\right]\right\rangle\)

for \(\begin{bmatrix} v_1 \\\ v_2 \\\ v_3 \end{bmatrix} ∈\) ℝ3

\(=\left\langle\left[\begin{array}{c} α v_1+v_2+v_3 \\ v_1 \\ v_1+β v_3 \end{array}\right],\left[\begin{array}{l} v_1 \\ v_2 \\ v_3 \end{array}\right]\right\rangle=\left[\begin{array}{ll} v_1 v_2 v_3 \end{array}\right]\left[\begin{array}{c} α v_1+v_2+v_3 \\ v_1 \\ v_1+β v_3 \end{array}\right]\)

= α\(v_1^2\) + v1v2 + v1v3 + v1v2 + v1v3 + β\(v_3^2\)

Take \(v = \begin{pmatrix} 1 \\\ 0 \\\ 0 \end{pmatrix}\) then α, β v, v > = α

Take α = - 1 and β = - 1 ⇒ α, β v, v > less than 0

So αβ > 0 but A is not positive

Option (2) is false.

α + β + 4 = 2 > 0 but A is not positive.

Option (3) is false.

(α, β) = (-1, -1) ∈ ℝ2 but A is not positive.

Option (4) is false.

hence option (1) is true.

Concept:

An inner product on V is a function satisfying the following conditions

(i) 1 + x₂, y> = 1, y> + <x2, y>

(ii) = c

(iii) <β, α> = \(<\overline{α, \beta}>\)

(iv) <α, α> = 0 if and only if α = 0 and <α, α> ≥ 0 for all α

Explanation:

Option (1): for c = - 1

〈cf, g〉 = 〈- f, g〉 = |\(\int_0^1\) - f(t) g(t) dt| = |\(\int_0^1\) f(t) g(t) dt|

and c〈f, g〉 = - 〈f, g〉 = - |\(\int_0^1\) f(t) g(t) dt|

Hence 〈cf, g〉 ≠ c〈f, g〉 

〈f, g〉 = |\(\int_0^1\) f(t) g(t) dt| is not inner product space.

Option (1) is false.

Option (2): 

for c = - 1

〈cf, g〉 = 〈- f, g〉 = \(\int_0^1\) -| f(t) g(t)| dt = \(\int_0^1\) |f(t) g(t)| dt

and c〈f, g〉 = - 〈f, g〉 = - \(\int_0^1\) |f(t) g(t)| dt

Hence 〈cf, g〉 ≠ c〈f, g〉 

〈f, g〉 = \(\int_0^1\) |f(t) g(t)| dt is not inner product space.

Option (2) is false

option (3): Let f(x) = x(x - 1)

then = f(0) f(0) + f(1)f(1)

= 0 + 0 = 0

but f(x) ≠ 0

so, = f(0) g(0) + f(1)g(1) is not inner product

Option (3) is false

option (4): = \(\int_0^1 f(t) g(t) dt\)

then

(1) 1 + f2, g> = \(\int_0^1 (f_1(t) + f_2(t)g(t)dt\)

\(= \int_0^1 (f_1(t) g(t)dt + \int_0^1 (f_2(t) g(t)dt\)

= 1, g> + 2, g>

(ii) = \(\int_0^1 cf(t) g(t)dt \) = \(c\int_0^1 f(t) g(t)dt \) = c

(iii) = \(​​\int_0^1 g(t) f(t) dt \) = \(​​\int_0^1 f(t) g(t) dt \) =

(iv) = \(​​\int_0^1 f(t) g(t) dt \) = \(​​\int_0^1 (f(t)) ^2dt \) ≥ 0 for all f

and \(​​\int_0^1 (f(t)) ^2dt \) = 0 ⇔ f = 0

and \(​​\int_0^1 (f(t)) ^2dt \) > 0 if f ≠ 0

So, = \(​​\int_0^1 f(t) g(t) dt \) satisfy all the properties

⇒ is an inner product

Option (4) is true.

Explanation:

(I) Let S = {(0, 1, 0), (0, 2, 0), (1, 0, 0)}, k = 3, n = 3

For linearly dependent,

C₁(0, 1, 0) + C2 (0, 2, 0) + C3 (1, 0, 0) = (0, 0, 0)

⇒ C3 = 0, C1 + 2C2 = 0 ⇒ C1 = -2C2

Here C1 = -2C2, C3 = 0 ⇒ not all Ci's are zero.

⇒ S is linearly dependent.

but here u3 can not be expressed as a linear combination of u1 & u2

Statement (I) false

(II) Take, k = 3, n = 3 and S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

Then S is Linearly independent.

but k \(\nless\) n

⇒ Statement (II) is false.

So, option (3) correct