Answer
$$
P^{-1}=\left[ \begin {array}{cccccc} 0&\frac{1}{3}&-\frac{1}{3}\\ -\frac{1}{9}&\frac {14}{27}&-\frac {2}{27}\\ \frac{2}{9}&-\frac{1}{27}&{\frac {4}{27}}\end {array} \right].
$$
Work Step by Step
Given
$$
B=\{(1,0,0),(0,1,0),(0,0,1)\}, \quad B^{\prime}=\{(2,-1,4),(0,2,1),(-3,2,1)\}.
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 2&-1&4&1&0&0\\ 0&2&1&0
&1&0\\ -3&2&1&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&0&\frac{1}{3}&-\frac{1}{3}\\ 0&
1&0&-\frac{1}{9}&\frac {14}{27}&-\frac {2}{27}\\ 0&0&1&\frac{2}{9}&-\frac{1}{27}&{\frac {4}{27}}\end {array} \right]
.$$
So, we have
$$
P^{-1}=\left[ \begin {array}{cccccc} 0&\frac{1}{3}&-\frac{1}{3}\\ -\frac{1}{9}&\frac {14}{27}&-\frac {2}{27}\\ \frac{2}{9}&-\frac{1}{27}&{\frac {4}{27}}\end {array} \right].
$$
