Answer
$$
P^{-1}=\left[ \begin {array}{cccccc} 1&3&2\\ 2
&-1&2\\ 5&6&1\end {array} \right]
$$
Work Step by Step
Given
$$
B=\{(1,3,2),(2,-1,2),(5,6,1)\}, \quad B^{\prime}=\{(1,0,0),(0,1,0),(0,0,1)\}.
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 1&0&0&1&3&2\\ \ 0&1&0&2
&-1&2\\ 0&0&1&5&6&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&1&3&2\\ \ 0&1&0&2
&-1&2\\ 0&0&1&5&6&1\end {array} \right]
.$$
So, we have
$$
P^{-1}=\left[ \begin {array}{cccccc} 1&3&2\\ 2
&-1&2\\ 5&6&1\end {array} \right]
$$
