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