Answer
$$
P^{-1}= \left[ \begin {array}{cccccccccc} 1&0&-\frac{5}{4}&-\frac{3}{4}&0
\\ -\frac{3}{11}&-\frac{2}{11}&{\frac {9}{22}}&\frac{1}{2}&-\frac{1}{11}
\\ {\frac {5}{11}}&{\frac {3}{22}}&-{
\frac {19}{44}}&-\frac{1}{4}&-\frac{2}{11}\\ 0&0&-\frac{1}{4}&\frac{1}{4}&0
\\ -{\frac {7}{11}}&-\frac{1}{11}&{\frac {21}{22}}
&\frac{1}{2}&{\frac {5}{11}}\end {array} \right].
$$
Work Step by Step
Given
$$
B=\{(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1) \} ,
\\ B^{\prime}= \{(1,2,4,-1,2),(-2,-3,4,2,1),(0,1,2,-2,1),(0,1,2,2,1),(1,-1,0,1,2) \} .
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccccc} 1&2&4&-1&2&1&0&0&0&0
\\ -2&-3&4&2&1&0&1&0&0&0\\ 0&1&2&-
2&1&0&0&1&0&0\\ 0&1&2&2&1&0&0&0&1&0
\\ 1&-1&0&1&2&0&0&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{5}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccccc} 1&0&0&0&0&1&0&-\frac{5}{4}&-\frac{3}{4}&0
\\ 0&1&0&0&0&-\frac{3}{11}&-\frac{2}{11}&{\frac {9}{22}}&\frac{1}{2}&-\frac{1}{11}
\\ 0&0&1&0&0&{\frac {5}{11}}&{\frac {3}{22}}&-{
\frac {19}{44}}&-\frac{1}{4}&-\frac{2}{11}\\ 0&0&0&1&0&0&0&-\frac{1}{4}&\frac{1}{4}&0
\\ 0&0&0&0&1&-{\frac {7}{11}}&-\frac{1}{11}&{\frac {21}{22}}
&\frac{1}{2}&{\frac {5}{11}}\end {array} \right]
.$$
So, we have
$$
P^{-1}= \left[ \begin {array}{cccccccccc} 1&0&-\frac{5}{4}&-\frac{3}{4}&0
\\ -\frac{3}{11}&-\frac{2}{11}&{\frac {9}{22}}&\frac{1}{2}&-\frac{1}{11}
\\ {\frac {5}{11}}&{\frac {3}{22}}&-{
\frac {19}{44}}&-\frac{1}{4}&-\frac{2}{11}\\ 0&0&-\frac{1}{4}&\frac{1}{4}&0
\\ -{\frac {7}{11}}&-\frac{1}{11}&{\frac {21}{22}}
&\frac{1}{2}&{\frac {5}{11}}\end {array} \right].
$$