Answer
$$
P^{-1}=\left[ \begin {array}{cccccc} {\frac {25}{11}}&{\frac {14}{11}}
&{\frac {16}{11}}\\ {\frac {9}{11}}&{\frac {9}
{11}}&\frac{4}{11}\\ \frac{1}{11}&\frac{1}{11}&{\frac {9}{11}}
\end {array} \right].
$$
Work Step by Step
Given
$$
B=\{(3,2,1),(1,1,2),(1,2,0)\}, \quad B^{\prime}=\{(1,1,-1),(0,1,2),(-1,4,0)\}.
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 1&1&-1&3&2&1\\ 0&1&2&
1&1&2\\ -1&4&0&1&2&0\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&{\frac {25}{11}}&{\frac {14}{11}}
&{\frac {16}{11}}\\ 0&1&0&{\frac {9}{11}}&{\frac {9}
{11}}&\frac{4}{11}\\ 0&0&1&\frac{1}{11}&\frac{1}{11}&{\frac {9}{11}}
\end {array} \right]
.$$
So, we have
$$
P^{-1}=\left[ \begin {array}{cccccc} {\frac {25}{11}}&{\frac {14}{11}}
&{\frac {16}{11}}\\ {\frac {9}{11}}&{\frac {9}
{11}}&\frac{4}{11}\\ \frac{1}{11}&\frac{1}{11}&{\frac {9}{11}}
\end {array} \right].
$$
