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Let $A=\left[\begin{array}{ll}0 & \alpha \\ 0 & 0\end{array}\right]$ and $(A+I)^{50}-50 A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$,
find $a b c+a b d+b c d+a c d$
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find $a b c+a b d+b c d+a c d$
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Verified Answer
The correct answer is:
0
As $\mathrm{A}^{2}=0, \mathrm{~A}^{k}=0 \forall k \geq 2$
Thus, $\quad(\mathrm{A}+\mathrm{I})^{50}=\mathrm{I}+50 \mathrm{~A} \Rightarrow(\mathrm{A}+\mathrm{I})^{50}-50 \mathrm{~A}=\mathrm{I}$
$\therefore \quad \mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=1$
$\mathrm{abc}+\mathrm{abd}+\mathrm{bed}+\mathrm{acd}=0$
Thus, $\quad(\mathrm{A}+\mathrm{I})^{50}=\mathrm{I}+50 \mathrm{~A} \Rightarrow(\mathrm{A}+\mathrm{I})^{50}-50 \mathrm{~A}=\mathrm{I}$
$\therefore \quad \mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=1$
$\mathrm{abc}+\mathrm{abd}+\mathrm{bed}+\mathrm{acd}=0$
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