\\
 &  & 0 \\
\end{array}\right).$$ Thus we may assume that $B_n$ has $\c{3}{2}=0$. Then by , $a_2$ or $a_3$ must be zero. Therefore we are in **Case 2** and the proposition is proved.

It remains to prove the claim. $$\begin{aligned}
B_3 &= P({\mathbb{C}}\oplus (\gamma_1^{c} \otimes \gamma_2^{-2c})) \\
&\cong P(({\mathbb{C}}\oplus (\gamma_1^{c} \otimes \gamma_2^{-2c}))\otimes \gamma_2^{c}) \\
&\cong P(\gamma_2^{c} \oplus (\gamma_1^{c} \otimes \gamma_2^{-c})). \\\end{aligned}$$ The total Chern class of $\gamma_2^{c} \oplus (\gamma_1^{c}\otimes\gamma_2^{-c})$ is $$c(\gamma_2^{c} \oplus (\gamma_1^{c} \otimes \gamma_2^{-c})) = (1+cx_2)(1+cx_1 - cx_2) = 1+ cx_1$$ since $x_2^2 = x_1x_2$ in $ H^4(B_2)$.

On the other hand, $c({\mathbb{C}}\oplus \gamma_1^c)=1+cx_1$. Therefore, $\gamma_2^{c} \oplus (\gamma_1^{c} \otimes \gamma_2^{-c}) \cong {\mathbb{C}}\oplus \gamma_1^{c}$ as bundles over $B_2$. Thus, $B_3 \cong P({\mathbb{C}}\oplus \gamma_1^c)= B_3'$ which has the associated matrix $$\left(
\begin{array}{ccc}
0 & 1 &c \\
 & 0 & 0 \\
 &  & 0 \\
\end{array}\right).$$

Now let $B_{n-1}\cong (\mathbb CP^1)^{n-1}$, and for $\alpha \in H^2(B_{n-1})$ let $\gamma^\alpha$ be the complex line bundle over $B_{n-1}$ with $c_1(\gamma^\alpha)=\alpha$ as before.

\[proposition:isomorphism of sum of two line bundle\] Let $\xi_1=\gamma^{\alpha_1} \oplus \gamma^{\alpha_2}$ and $\xi_2=\gamma^{\beta_1} \oplus \gamma^{\beta_2}$ be sums of two line bundles over $B_{n-1}\cong (\mathbb CP^1)^{n-1}$ such that $c_1(\xi_1)=c_1(\xi_2)$ and $c_2(\xi_1)=c_2(\xi_2)=0$. Then $\xi_1$ and $\xi_2$ are isomorphic.

Let $H^\ast(B_{n-1})\cong \mathbb Z[x_1,\ldots, 