## Elementary Linear Algebra 7th Edition

$$P^{-1}=\left[ \begin {array}{cccc} \frac{3}{2}&-2\\ -\frac{1}{2}&1 \end {array} \right] .$$
Given $$B=\{(1,0),(0,1)\}, B^{\prime}=\{(2,4),(1,3)\} .$$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$\left[B^{\prime} B\right]=\left[ \begin {array}{cccc} 2&4&1&0\\ 1&3&0&1 \end {array} \right] .$$ Using Gauss-Jordan elimination, the reduced row echelon form is $$\left[\begin{array}{ll}{I_{2}} & {P^{-1}}\end{array}\right]=\left[ \begin {array}{cccc} 1&0&\frac{3}{2}&-2\\ 0&1&-\frac{1}{2}&1 \end {array} \right] .$$ So, you have $$Given$$ B=\{(1,0),(0,1)\}, B^{\prime}=\{(2,4),(1,3)\} .$$To find the transition matrix from B to B^{\prime}, we form the matrix$$ \left[B^{\prime} B\right]=\left[ \begin {array}{cccc} 2&4&1&0\\ 1&3&0&1 \end {array} \right] .$$Using Gauss-Jordan elimination, the reduced row echelon form is$$ \left[\begin{array}{ll}{I_{2}} & {P^{-1}}\end{array}\right]=\left[ \begin {array}{cccc} 1&0&\frac{3}{2}&-2\\ 0&1&-\frac{1}{2}&1 \end {array} \right] .$$So, you have$$ P^{-1}=\left[ \begin {array}{cccc} \frac{3}{2}&-2\\ -\frac{1}{2}&1 \end {array} \right] .