Linear Algebra MCQs and MSQs with Solutions for CSIR NET, IIT JAM : Linear Transformation - V

41. Let the linear transformations \(S\) and \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be defined by

\(S(x, y, z)=(2 x, 4 x-y, 2 x+3 y-z)\)

\(T(x, y, z)=(x \cos \theta-y \sin \theta,{x}{\sin } \theta+y \cos \theta, z)\) where \(0<\theta<\pi / 2\).

Then

(a.) \(S\) is one to one but not \(T\)
(b.) \(T\) is one to one but not \(S\)
(c.) Both \(S\) and \(T\) are one to one
(d.) Neither \(S\) nor \(T\) is one to one

Explanation for Question 41:
Observe co-domain of \(T=\mathbb{R}^{3} \Rightarrow \theta\) is fixed. Now \(S=\left[\begin{array}{ccc}2 & 0 & 0 \\ 4 & -1 & 0 \\ 2 & 3 & -1\end{array}\right] \quad \& \quad T=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\) \[ |S|=2 \neq 0 \ \& \ |T|=1 \neq 0 \] \(\Rightarrow S \& T\) are invertible. This completes the explanation for Question 41.

42. Consider the vector space \(\mathbb{R}^{3}\) and the maps \(f, g: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) defined by \(f(x, y, z)=(x,|y|, z)\) and \(g(x, y, z)=(x+1, y-1, z)\). Then

(a.) Both \(f\) and \(g\) are linear
(b.) Neither \(f\) nor \(g\) is linear
(c.) \(g\) is linear but not \(f\)
(d.) \(f\) is linear but not \(g\)

Explanation for Question 42:
\[ f(0,1,0)=(0,1,0)=f(0,-1,0) \] \[ \begin{aligned} & f(0,0,0)=f(0,1,0)+f(0,-1,0)=(0,2,0) \\ & \Rightarrow f(0,0,0)=(0,2,0) \neq(0,0,0) \end{aligned} \] \(\Rightarrow f\) is not a linear transformation. \[ g(0,0,0)=(1,-1,0) \neq(0,0,0) \] \(\Rightarrow g\) is not a linear transformation. This completes the explanation for Question 42.

43. Let \(S\) and \(T\) be two linear operators on \(\mathbb{R}^{3}\) defined by \(S(x, y, z)=(x, x+y, x-y-z)\) and \(T(x, y, z)=(x+2 z, y-z, x+y+z)\). Then

(a.) \(S\) is invertible but not \(T\)
(b.) \(T\) is invertible but not \(S\)
(c.) Both \(S\) and \(T\) are invertible
(d.) Neither \(S\) nor \(T\) is invertible

Explanation for Question 43:
\[ [S]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & -1 & -1\end{array}\right] \quad \& \quad [T]=\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 1 & -1 \\ 1 & 1 & 1\end{array}\right] \] \(\Rightarrow \operatorname{Rank}(S)=3\) & Rank \((T)=2\) This completes the explanation for Question 43.

44. Let \(\mathrm{V}, \mathrm{W}\) and \(\mathrm{X}\) be three finite dimensional vector spaces such that \(\operatorname{dim} V=\operatorname{dim} X\). Suppose \(S: V \rightarrow W\) and \(T: W \rightarrow X\) are two linear maps such that \(\operatorname{ToS}: V \rightarrow X\) is injective. Then

(a.) \(\mathrm{S}\) and \(\mathrm{T}\) are surjective
(b.) \(\mathrm{S}\) is surjective and \(\mathrm{T}\) is injective
(c.) \(\mathrm{S}\) and \(\mathrm{T}\) are injective
(d.) \(\mathrm{S}\) is injective and \(\mathrm{T}\) is surjective

Explanation for Question 44:
For any well-defined function, not necessarily linear, \(f\) and \(g\), if \(f \circ g\) is \(1-1\), then \(g\) is \(1-1\). If \(f \circ g\) is onto, then \(f\) is onto. Here, \(ToS: V \rightarrow X\) is \(1-1\) with \(\operatorname{dim} V=\operatorname{dim} X\) \(\Rightarrow T \circ S\) is bijective. This completes the explanation for Question 44.

45. Let \(\mathbb{R}^{2 \times 2}\) be the real vector space of all \(2 \times 2\) real matrices. For \(Q=\left(\begin{array}{cc}1 & -2 \\ -2 & 4\end{array}\right)\), define a linear transformation \(T\) on \(\mathbb{R}^{2 \times 2}\) as \(T(P)=Q P\). Then the rank of \(T\) is

(a.) 1
(b.) 2
(c.) 3
(d.) 4

Explanation for Question 45:
Let \(B=\left\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\}\) be the basis of \(\mathbb{R}^{2 \times 2}\) \(T: \mathbb{R}^{2 \times 2} \longrightarrow \mathbb{R}^{2 \times 2}\) is defined as \(T(P)=Q P\) where \(Q=\left[\begin{array}{cc}1 & -2 \\ -2 & 4\end{array}\right]\) (\(\operatorname{rank}(Q)=1\)) \[ \Rightarrow[T]_{B}^{B}=\left[\begin{array}{cccc} 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -2 \\ -2 & 0 & 4 & 0 \\ 0 & -2 & 0 & 4 \end{array}\right]_{4 \times 4} \] \(\operatorname{Rank} T=2\) This completes the explanation for Question 45.

46. Let \(\mathrm{T}\) be an arbitrary linear transformation from \(\mathbb{R}^{n}\) to \(\mathbb{R}^{n}\) which is not one-one.

Then

(a.) Rank \(\mathrm{T}>0\)
(b.) Rank \(\mathrm{T}=\mathrm{n}\)
(c.) Rank \(\mathrm{T}<\mathrm{n}\)
(d.) Rank \(\mathrm{T}=\mathrm{n}-1\)

Explanation for Question 46:
\(T\) is not \(1-1 \Rightarrow\) two linearly independent vectors have the same image. \(\Rightarrow\) Rank \(T < n\) but not necessarily \(n-1\). This completes the explanation for Question 46.

47. Let \(\mathrm{T}\) be a linear transformation from \(\mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) defined by \(\mathrm{T}(x, y, z)=(x+y, y-z)\). Then the matrix of \(\mathrm{T}\) with respect to the ordered bases \(\{(1,1,1),(1,-1,0),(0,1,0)\}\) and \(\{(1,1),(1,0)\}\) is

(a.) \(\left[\begin{array}{ccc}-2 & 0 & 1 \\ 1 & 1 & -1\end{array}\right]\)
(b.) \(\left[\begin{array}{ccc}0 & -1 & 1 \\ 2 & 1 & 0\end{array}\right]\)
(c.) \(\left[\begin{array}{cc}2 & 1 \\ 0 & -1 \\ 1 & 1\end{array}\right]\)
(d.) \(\left[\begin{array}{cc}0 & 2 \\ -1 & 1 \\ 1 & 0\end{array}\right]\)

Explanation for Question 47:
\[ \begin{aligned} & T(x, y, z)=(x+y, y-z) \\ & \Rightarrow T(1,1,1)=(2,0)=2 e_{2} \\ & T(1,-1,0)=(0,-1)=-e_{1}+e_{2} \\ & T(0,1,0)=(1,1)=e_{1} \\ & \Rightarrow[T]=\left[\begin{array}{lll} 0 & -1 & 1 \\ 2 & 1 & 0 \end{array}\right] \end{aligned} \] This completes the explanation for Question 47.

48. Choose the correct matching from A, B, C, and \(\mathrm{D}\) for the transformation \(T_{1}, T_{2}\) and \(T_{3}\) (mapping from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{3}\)) as defined in Group 1 with the statements given in Group 2.
Group 1
P. \(T_{1}(x, y)=(x, x, 0)\)
Q \(T_{2}(x, y)=(x, x+y, y)\)
R \(T_{3}(x, y)=(x, x+1, y)\)
Group 2 :
1. Linear transformation of rank 2
2. Not a linear transformation
3. Linear transformation

(a.) P-3, Q-1, R-2
(b.) P-1, Q-2, R-3
(c.) P-3, Q-2, R-1
(d.) P-1, Q-3, R-2

Explanation for Question 48:
\[ T_{3}(x, y)=(x, x+1, y) \] is not a linear transformation as \(T_{3}(0,0) \neq(0,0,0)\). \[ T_{1}(x, y)=(x, x, 0) \] is a linear transformation but of rank 1. \(\Rightarrow R-2, P-3, Q-1\) is the correct combination. This completes the explanation for Question 48.

49. Consider the basis \(S=\left\{v_{1}, v_{2}, v_{3}\right\}\) for \(\mathrm{R}^{3}\) where \(v_{1}=(1,1,1), V_{2}=(1,1,0), v_{3}=(1,0,0)\) and let \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) be a linear transformation such that \(T v_{1}=(1,0), T v_{2}=(2,-1), T v_{3}=(4,3)\). Then \(T(2,-3,5)\) is

(a.) (-1,5)
(b.) (3,4)
(c.) (0,0)
(d.) (9,23)

Explanation for Question 49:
\[ \begin{aligned} (2,-3,5) & =5 v_{1}-8 v_{2}+5 v_{3}^{2} \\ \Rightarrow T(2,-3,5) & =5 T\left(v_{2}\right)-8 T\left(v_{2}\right)+5 T\left(v_{3}\right) \\ & =5(1,0)-8(2,-1)+5(4,3) \\ & =(9,23) \end{aligned} \] Note: when the basis is not given, we assume the standard basis. This completes the explanation for Question 49.

50. For a positive integer \(n\), let \(P_{n}\) denote the vector space of polynomials in one variable \(x\) with real coefficients and with degree \(\leq n\). Consider the map \(T: P_{2} \rightarrow P_{4}\) defined by \(T(p(x))=p\left(x^{2}\right)\). Then

(a.) \(T\) is a linear transformation and \(\operatorname{dim}\) range \((T)=5\)
(b.) \(T\) is a linear transformation and dim range \((T)=3\)
(c.) \(T\) is a linear transformation and dim range \((T)=2\)
(d.) \(T\) is not a linear transformation.

Explanation for Question 50:
Make the matrix with standard basis. You will get it of full rank.