Let $T^{(i)}=A^{i-1}b$ for $i=1,\cdots n$. You can solve $\widehat{b}$ from the equation $T\widehat{b}=b$. By Cramer's rule,

$$ \widehat{b_1}=\frac{det[b, T^{(2)}, \cdots, T^{(n)}]} {det T} $$ $$ \widehat{b_2}=\frac{det[b, b, T^{(3)}, \cdots, T^{(n)}]} {det T} $$ $\cdots$ $$ \widehat{b_n}=\frac{det[b, T^{(2)}, \cdots, T^{(n-1)},b]} {det T} $$ Then you can see that $\widehat{b_1}=1, \widehat{b_i}=0$ for all $i=2,\cdots, n$.

Let $T^{(i)}=A^{i-1}b$ for $i=1,\cdots n$. You can solve $\widehat{b}$ from the equation $T\widehat{b}=b$. By Cramer's rule,

$$ \widehat{b_1}=\frac{det[b, T^{(2)}, \cdots, T^{(n)}]} {det T} $$ $$ \widehat{b_2}=\frac{det[b, b, T^{(3)}, \cdots, T^{(n)}]} {det T} $$ $\cdots$ $$ \widehat{b_n}=\frac{det[b, T^{(2)}, \cdots, T^{(n-1)},b]} {det T} $$ Then you can see that $\widehat{b_1}=1, \widehat{b_i}=0$ for all $i=2,\cdots, n$.

Further for $\widehat{A}$, use Cramer's rule to each column vector of $\widehat{A}$, in the equation $T\widehat{A} = AT$.