Question

Assertion :$A$ is the $n\times n$ matrix whose elements are all ${}^{\prime }{1}^{\prime }$ and $B$ is the $n\times n$ matrix whose diagonal elements are all $n$ and other elements are ${}^{\prime }n-r^{\prime }$> then $(B-rI)[B-(n^2-nr+r\left)I\right]=0$ because Reason: $A^2$ is a scalar multiple of $A$.

• 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. Both Assertion and Reason are incorrect

Answer 1

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

Hree $A\left[\begin{array}{ccc}1& 1& ...1\\ 1& 1& ...1\\ 1& 1& ...1\end{array}\right]$

$\therefore A^2=A.A=\left[\begin{array}{ccc}1& 1& ...1\\ 1& 1& ...1\\ 1& 1& ...1\end{array}\right]\left[\begin{array}{ccc}1& 1& ...1\\ 1& 1& ...1\\ 1& 1& ...1\end{array}\right]$

$=\left[\begin{array}{ccc}n& n& ...n\\ n& n& ...n\\ ...& ...& ...n\\ n& n& ...n\end{array}\right]=nA$

$B=\left[\begin{array}{ccc}n& n-r& ...n-r\\ n-r& n& ...n-r\\ ...& ...& ...\\ n-r& n-r& ...n\end{array}\right]$

$\therefore B-rI=\left[\begin{array}{ccc}n-r& n-r& ...n-r\\ n-r& n-r& ...n-r\\ ...& ...& ...\\ n-r& n-r& ...n-r\end{array}\right]$

$=(n-r)A$

Hence, $(B-rI)[B-(n^2-nr+r)I]$

$=(B-rI)[(B-rI)-n(n-r)I]$

$=(n-r)A[(n-r)A-n(n-r)I]$

$={(n-r)}^{2}A(A-nI)$

$={(n-r)}^2{A}^2-n{(n-r)}^{2}AI$

$={(n-r)}^2[A^2-nA]$

$={(n-r)}^2[nA-nA][\because A^2=nA]$

$={(n-r)}^2\left(0\right)=0$