Matrices and Determinants 2 Question 22
24. Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N=N M$. Further, if $M \neq N^{2}$ and $M^{2}=N^{4}$, then
(2014 Adv.)
(a) determinant of $\left(M^{2}+M N^{2}\right)$ is 0
(b) there is a $3 \times 3$ non-zero matrix $U$ such that $\left(M^{2}+M N^{2}\right) U$ is zero matrix
(c) determinant of $\left(M^{2}+M N^{2}\right) \geq 1$
(d) for a $3 \times 3$ matrix $U$, if $\left(M^{2}+M N^{2}\right) U$ equals the zero matrix, then $U$ is the zero matrix
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Answer:
Correct Answer: 24. (a, b)
Solution:
- PLAN: (i) If $A$ and $B$ are two non-zero matrices and $A B=B A$, then $(A-B)(A+B)=A^{2}-B^{2}$
(ii) The determinant of the product of the matrices is equal to product of their individual determinants, i.e. $|A B|=|A||B|$.
Given, $M^{2}=N^{4} \Rightarrow M^{2}-N^{4}=0$
$ \Rightarrow \quad\left(M-N^{2}\right) \quad\left(M+N^{2}\right)=0 $
[as $M N=N M$ ]
Also, $M \neq N^{2}$
$ \Rightarrow \quad M+N^{2}=0 $
$\Rightarrow \quad \operatorname{det}\left(M+N^{2}\right)=0$
Also, $\operatorname{det}\left(M^{2}+M N^{2}\right)=(\operatorname{det} M)\left(\operatorname{det} M+N^{2}\right)$
$ =(\operatorname{det} M)(0)=0 $
As, $\quad \operatorname{det}\left(M^{2}+M N^{2}\right)=0$
Thus, there exists a non-zero matrix $U$ such that
$ \left(M^{2}+M N^{2}\right) U=0 $
