Question

Assertion :If $A=\left[\begin{array}{cc}1& \pi \\ 0& 1\end{array}\right]$, then $A^{100}=\left[\begin{array}{cc}1& 100\pi \\ 0& 1\end{array}\right]$. Reason: If $B$ is a $2\times 2$ matrix such that $B^2=0$, then $(I+B{)}^n=I+nB$ for each $n\in N$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Since $B$ commutes with $I$, we can use binomial theorem to obtain

$(I+B{)}^n=I+\left[\begin{array}{c}n\\ 1\end{array}\right]B+\left[\begin{array}{c}n\\ 1\end{array}\right]B^2+...+\left[\begin{array}{c}n\\ n\end{array}\right]B"$
Since $B^2=0$, we get $B^r=0\forall r\ge 2$
Thus,
$(I+B{)}^n=I+nB$
Now,
$A=I+B$ where $B=\left[\begin{array}{cc}0& \pi \\ 0& 0\end{array}\right]$
Since $B^2=0$, we get
$A^{100}=I+100B=\left[\begin{array}{cc}1& 100\pi \\ 0& 1\end{array}\right]$
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.