Answer
$$
P^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ 2
&\frac{1}{2}&-\frac{1}{3}\\ -\frac{1}{2}&0&\frac{1}{2}\end {array} \right]
$$
Work Step by Step
Given
$$
B=\{(1,0,0),(0,1,0),(0,0,1)\} , \quad B^{\prime}=\{(1,0,0),(0,2,8),(6,0,12)\}
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 0&2&8&0
&1&0\\ 6&0&12&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 0&1&0&2
&\frac{1}{2}&-\frac{1}{3}\\ 0&0&1&-\frac{1}{2}&0&\frac{1}{2}\end {array} \right]
.$$
So, we have
$$
P^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ 2
&\frac{1}{2}&-\frac{1}{3}\\ -\frac{1}{2}&0&\frac{1}{2}\end {array} \right]
$$
