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:

  1. 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 $



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