Inverse of a matrix using bases Consider the bases 
$(4,2)$ , $(5,3)$ and $(1,0)$,$(0,1)$ of $F^2$. The Matrix of 
$$
M(I,\bigg(\Big(4,2),(5,3)),((1,0),(0,1)\Big)\bigg) = \Bigg(\begin{matrix}4 &5 \\2 &3 \end{matrix}\Bigg)
$$
The matrix above was computed using the procedure 
$I(4,2) = 4(1,0) + 2(0,1)$ and 
$I(5,3) = 5(1,0) + 5(0,1)$.
but when I tried finding the matrix of the reverse order of bases, I had
$$
M(I,\bigg(\Big(1,0),(0,1)),((4,2),(5,3)\Big)\bigg) = \Bigg(\begin{matrix}4 &5 \\2 &3 \end{matrix}\Bigg)
$$
I computed the matrix of the reverse order of the bases in the form below.
$I(1,0) = 1(4,2) + 0(5,3)$ and 
$I(0,1) = 0(4,2) + 1(5,3)$.
But the reverse order did not give me the inverse matrix. 
Please how was the inverse calculated ?

You calculated the first matrix correctly by writing $(4,2)$ and $(5,3)$ as linear combinations of $(1,0)$ and $(0,1)$.  You should calculate the inverse matrix by writing $(1,0)$ and $(0,1)$ as linear combinations of $(4,2)$ and $(5,3)$:
$$(1,0) = \frac{3}{2}(4,2) + (-1) (5,3)$$
$$(0,1) = -\frac{5}{2}(4,2) + 2 (5,3)$$
The inverse matrix is therefore
$$\begin{pmatrix} \frac{3}{2} & - \frac{5}{2} \\ -1 & 2 \end{pmatrix}$$
