Let $T$ be the linear operator on $\mathbb R^3$ defined on $$(_1, _2, _3 ) = (3_1 + _3, −2_1 + _2, −_1 + 2_2 + 4_3 )$$
Find the transition matrix $P$ from the ordered standard basis of $\mathbb R^3$ to the ordered basis $′ = {(1,0,1), (−1, 2, 1), (2,1,1)}.\tag*{}$ Hence find the matrix of $$ relative to ordered basis $′$.
$$\text{Sol}^{\text{n}}.:-$$ $$[T]_{B'}=P^{-1}[T]_B P$$ I have found $$[T]_{B}= \begin{pmatrix} 3 & 0 & 1 \\ -2 & 1 & 0 \\ -1 & 2 & 4 \\ \end{pmatrix}$$ How to find $P$? Is there another method for finding $[T]_{B'}$ without finding $P$?
Matrix of a linear transformation
Let:
Each $v_i\in B$ is mapped to some vector in $W$.
$T(v_i)\in W$ so it can be expressed as a unique linear combination of vectors in $B'$.
Suppose each $\begin{align}T(v_i)&=\lambda_{1i} w_1+\lambda_{2i} w_2+\ldots+\lambda_{mi}w_m\\ &=[\lambda_{1i},\lambda_{2i},\ldots,\lambda_{mi}]_{B'}\end{align}\tag*{}$
Matrix of $T$ is given by putting these coordinates into columns \begin{pmatrix} \lambda _{11} & \lambda _{12} & \dotsc & \lambda _{1n}\\ \lambda _{21} & \lambda _{22} & \dotsc & \lambda _{2n}\\ \vdots & \vdots & \ddots & \vdots \\ \lambda _{m1} & \lambda _{m2} & \dotsc & \lambda _{mn} \end{pmatrix}
such that each column $C_i$ corresponds to the coordinates of $T(v_i)$.
The matrix of $T$ depends on choice of bases of the domain and co-domain, so it's useful to denote this particular matrix by $[T]_{B'\leftarrow B}$.
This matrix, when right multiplied by a column of coordinates, finds the corresponding vector $v$ w.r.t. $B$ and spits out the transformed vector $T(v)$ in terms of coordinates w.r.t. $B'$.
For example, $[T]_{B'\leftarrow B}$ when right multiplied by $\begin{pmatrix}1\\ 0\\ \vdots\\ 0\end{pmatrix}$, the corresponding vector being $v_1$, it is mapped to $T(v_1)$ which is $\begin{pmatrix}\lambda_{11}\\ \lambda_{21}\\ \vdots\\ \lambda_{m1}\end{pmatrix}$ w.r.t. $B'$.
Let
Minding that two matrices are equal if they give the same output every time when right multiplied by any given column of coordinates, we can make these manipulations:
$\begin{align}& \ [T]_{X' \leftarrow X}\\&=[I]_{X' \leftarrow B'}\cdot[T]_{B' \leftarrow X}\\ &= [I]_{X' \leftarrow B'}\cdot[T]_{B' \leftarrow B}\cdot [I]_{B\leftarrow X}\end{align}$
where $I$ is understood to be the identity transformation (determine the domain and co-domain by the bases in the subscript). For instance, $[I]_{B\leftarrow X}$ is a matrix of the identity mapping on $V$ which when right multiplied by a column of coordinates, finds the corresponding vector w.r.t. $B$ and spits out the same in coordinates w.r.t. $X$.
Transition matrix
Definition: Let $V$ be an $n$-dimensional vector space having ordered bases $B=\{v_1, \ldots, v_n\}$ and $B'=\{v_1',\ldots, v_n'\}$ over a field $F$. There is a unique bijective linear transformation $\mathscr P:V\to V$ which maps each $v_i$ from $B$ to corresponding $v_i'$ in $B'$. The matrix of $\mathscr P$ w.r.t. $B$ i.e., $P=[\mathscr P]_{B}$ is called the transition matrix from $B$ to $B'$.
$\big([\mathscr P]_B$ is a simplified notation for $[\mathscr P]_{B\leftarrow B}$. $\big)$
Property: $\mathscr P$ “preserves” coordinates. Given a vector with coordinates $a_1,\ldots, a_n$ w.r.t. $B$, it will mapped to a vector with the same coordinates w.r.t. $B'$. $\displaystyle\mathscr P\left(\sum_{i=1}^n a_iv_i\right)=\sum_{i=1}^n a_iv_i'\tag*{}$
Given two linear transformations $V\to V$, their matrices are identical if and only if they map each corresponding basis vectors to the same coordinates.
Let $T:V\to V$ be an L.T.
Pick an arbitrary $v_i'$ from $B'$, corresponding to which, we have $v_i$ in $B$. Suppose $T(v_i')$ has coordinates $r_1,\ldots, r_n$ w.r.t. $B'$ i.e., $\displaystyle \begin{align} & T(v_i')=\sum_{j=1}^n r_j v_j'\\ \implies & T\circ \mathscr P(v_i) = \mathscr P\left(\sum_{j=1}^n r_jv_j\right)\\ \implies & \mathscr P^{-1}\circ T\circ \mathscr P(v_i)=\sum_{j=1}^n r_jv_j\end{align} \tag*{}$
$ \begin{align}\therefore [T]_{B'}&=\left[\mathscr P^{-1}\circ T\circ \mathscr P\right]_B \\ &= P^{-1}[T]_B P\end{align}$
Alternately,
$ \begin{align}[T]_{B'}&=[T]_{B'\leftarrow B'}\\&= [I]_{B' \leftarrow B}\cdot[T]_{B\leftarrow B}\cdot[I]_{B\leftarrow B'}\\ &= P^{-1}[T]_B P\end{align}$
I leave it as an exercise to prove that $P=[I]_{B\leftarrow B'}$.
Let's keep to the original post.
The bases: $B=\{(1,0,0), (0,1,0), (0,0,1)\}\tag*{}$ & $B'=\{(1,0,1),(-1,2,1),(2,1,1)\}\tag*{}$
Consider the L.T. $\mathscr P: \mathbb R^3\to \mathbb R^3$ which maps vectors in $B$ to corresponding vectors in $B'$.$$\mathscr P (1,0,0)=(1,0,1)\\ \mathscr P(0,1,0)=(-1,2,1)\\ \mathscr P(0,0,1)=(2,1,1)$$
The transition matrix $P$ is given by $$P=[\mathscr P]_B=\begin{pmatrix} 1 & -1 & 2\\ 0 & 2 & 1\\ 1 & 1 & 1 \end{pmatrix}$$
Alternately, one can find the transition matrix $P$ by finding $[I]_{B \leftarrow B'}$ where $I$ is the identity mapping on $\mathbb R^{3}$. (I leave that as an exercise.)
Given the definition of $T:\mathbb R^3\to \mathbb R^3$ in your post and that you have already found $[T]_{B}= \begin{pmatrix} 3 & 0 & 1 \\ -2 & 1 & 0 \\ -1 & 2 & 4 \\ \end{pmatrix},\tag*{}$ $[T]_{B'}$ can be computed as follows: $$\begin{align}[T]_{B'}&=P^{-1}[T]_BP\\&=\frac{1}{4}\begin{pmatrix} 17 & 35 & 22\\ -3 & 15 & -6\\ -2 & -14 & 0 \end{pmatrix}\end{align}$$
Here's an alternate way to find $[T]_{B'}$.
By definition, $[T]_{B'}=[T]_{B'\leftarrow B'}$.
Find solutions to $$T(1,0,1)=[\cdot, \cdot, \cdot]_{B'} \\T(-1,2,1)=[\cdot, \cdot, \cdot]_{B'}\\ T(2,1,1)=[\cdot, \cdot, \cdot]_{B'}$$ and put these coordinates into column...
$$\begin{align}& T(1,0,1)\\&=(4,-2,3)\\ &= a_{11}(1,0,1)+a_{21}(-1,2,1)+a_{31}(2,1,1)\\ &=[a_{11}, a_{21}, a_{31}]_{B'}\end{align}$$
Solve the system of linear equations: $\begin{pmatrix}1 & -1 & 2\\ 0&2&1\\ 1&1&1\end{pmatrix}\begin{pmatrix}a_{11}\\ a_{21}\\ a_{31}\end{pmatrix}=\begin{pmatrix}4\\-2\\3\end{pmatrix}\tag*{}$ and find $a_{11}=17/4$, $a_{21}=-3/4$ and $a_{31}=-2/4$.
Likewise, do the rest.
$\begin{pmatrix}1 & -1 & 2\\ 0&2&1\\ 1&1&1\end{pmatrix}\begin{pmatrix}a_{12}\\ a_{22}\\ a_{32}\end{pmatrix}=T(-1,2,1)\tag*{}$
$\begin{pmatrix}1 & -1 & 2\\ 0&2&1\\ 1&1&1\end{pmatrix}\begin{pmatrix}a_{13}\\ a_{23}\\ a_{33}\end{pmatrix}=T(2,1,1)\tag*{}$
It's tedious, good luck!
Give me feedback so that I can make this answer more understandable. I am constantly learning and updating this post according to my perspective.