Answer
$$
P^{-1}= \left[ \begin {array}{cccccccccc} {\frac {12}{157}}&{\frac
{45}{157}}&-{\frac {17}{157}}&-{\frac {1}{157}}&-{\frac {4}{157}}
\\ {\frac {32}{157}}&-{\frac {37}{157}}&{
\frac {7}{157}}&{\frac {47}{314}}&{\frac {31}{314}}
\\ {\frac {5}{314}}&-{\frac {99}{314}}&{
\frac {3}{157}}&{\frac {287}{628}}&{\frac {49}{628}}
\\ {\frac {10}{157}}&-{\frac {41}{157}}&{
\frac {12}{157}}&{\frac {103}{314}}&-{\frac {59}{314}}
\\ -{\frac {7}{157}}&{\frac {13}{157}}&{
\frac {23}{157}}&-{\frac {25}{314}}&{\frac {57}{314}}\end {array}\right]
.
$$
Work Step by Step
Given
$$
B=\{(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1) \} ,
\\ B^{\prime}= \{(2,4,-2,1,0),(3,-1,0,1,2),(0,0,-2,4,5), (2,-1,2,1,1),(0,1,2,-3,1) \} .
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccccc} 2&4&-2&1&0&1&0&0&0&0
\\ 3&-1&0&1&2&0&1&0&0&0\\ 0&0&-2&4
&5&0&0&1&0&0\\ 2&-1&2&1&1&0&0&0&1&0
\\ 0&1&2&-3&1&0&0&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{5}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccccc} 1&0&0&0&0&{\frac {12}{157}}&{\frac
{45}{157}}&-{\frac {17}{157}}&-{\frac {1}{157}}&-{\frac {4}{157}}
\\ 0&1&0&0&0&{\frac {32}{157}}&-{\frac {37}{157}}&{
\frac {7}{157}}&{\frac {47}{314}}&{\frac {31}{314}}
\\ 0&0&1&0&0&{\frac {5}{314}}&-{\frac {99}{314}}&{
\frac {3}{157}}&{\frac {287}{628}}&{\frac {49}{628}}
\\ 0&0&0&1&0&{\frac {10}{157}}&-{\frac {41}{157}}&{
\frac {12}{157}}&{\frac {103}{314}}&-{\frac {59}{314}}
\\ 0&0&0&0&1&-{\frac {7}{157}}&{\frac {13}{157}}&{
\frac {23}{157}}&-{\frac {25}{314}}&{\frac {57}{314}}\end {array}
\right]
.$$
So, we have
$$
P^{-1}= \left[ \begin {array}{cccccccccc} {\frac {12}{157}}&{\frac
{45}{157}}&-{\frac {17}{157}}&-{\frac {1}{157}}&-{\frac {4}{157}}
\\ {\frac {32}{157}}&-{\frac {37}{157}}&{
\frac {7}{157}}&{\frac {47}{314}}&{\frac {31}{314}}
\\ {\frac {5}{314}}&-{\frac {99}{314}}&{
\frac {3}{157}}&{\frac {287}{628}}&{\frac {49}{628}}
\\ {\frac {10}{157}}&-{\frac {41}{157}}&{
\frac {12}{157}}&{\frac {103}{314}}&-{\frac {59}{314}}
\\ -{\frac {7}{157}}&{\frac {13}{157}}&{
\frac {23}{157}}&-{\frac {25}{314}}&{\frac {57}{314}}\end {array}\right]
.
$$
