Let T:P3(R)→M2×2(R) be a linear transformation, then find the matrix of transformation with respect to...

Step 1
Let ${\displaystyle T:P_3\left(R\right)\to M_{2\times 2}\left(R\right)}$ be a linear transformation
To find:
(a) The matrix transformtion with respect to standard basis.
(b) The condition number of the matrix of transformation with
respect to $|||{|}_1$.
Step 2
(a) Consider a given transformation:
${\displaystyle T:P_3\left(R\right)\to M_{2\times 2}\left(R\right)bealineartransformation}$
Now, define a transformation:
$T(a+bx+c{x}^2+d{x}^3)=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Tye standard basis of vector space ${\displaystyle P_3\left(R\right)\text{and}{M}_{2time2}\left(R\right)}$ are
$B=\{a+bx+c{x}^2+d{x}^3\}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Further, we have
$T(1)=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]=1+(0)x+(0)x^2+(0)x^3$
$T(x)=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=(0)+(1)x+(0)x^2+(0)x^3$
$T(x^2)=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=(0)+(0)x+(1)x^2+(0)x^3$
$T(x^3)=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=(0)+(0)x+(0)x^2+(1)x^3$
Now, the matrix transformation w.r.t to the standard basis B is
$T_B^{B^{\prime }}\sim A=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

(b)
From a part (a):
$A=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]and{A}^{-1}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$
Now, the condition number of the matrix A w.r.t norm ${\displaystyle {\left|\left|\left|\left|{}_{\left\{1\right\}}is\right|\right|A\right|\right|}_1\cdot {\left|\left|A^{-1}\right|\right|}_1}$.
${\displaystyle {\left|\left|A\right|\right|}_1=\sqrt{1+1+1+1}=2\text{and}{\left|\left|A^{-1}\right|\right|}_1=sgrt\{1+1+1+1\}=2}$.
Further solving, we have
Condition number ${\displaystyle ={\left|\left|A^{-1}\right|\right|}_1\cdot {\left|\left|A^{-1}\right|\right|}_1=\left(2\right)\cdot \left(2\right)=4}$.
${\displaystyle \Rightarrow Conditionnumber=4}$.