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

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

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

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

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

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

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

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

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

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

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