Answer
$$
P^{-1}=\left[ \begin {array}{cccccc} 1&-3&3\\
1&2&-3\\ -1&-1&2\end {array} \right].
$$
Work Step by Step
Given
$$
B=\{(1,0,0),(0,1,0),(0,0,1)\}, \quad B^{\prime}=\{(1,3,3),(1,5,6),(1,4,5)\}.
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 1&3&3&1&0&0\\ 1&5&6&0
&1&0\\ 1&4&5&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&1&-3&3\\ 0&1&0&
1&2&-3\\ 0&0&1&-1&-1&2\end {array} \right]
.$$
So, we have
$$
P^{-1}=\left[ \begin {array}{cccccc} 1&-3&3\\
1&2&-3\\ -1&-1&2\end {array} \right].
$$
