Let M and N be two 3 ×3 matrices such that MN = NM. Further, if M ≠N2 and M2=N4 , then

(B) There is a 3 ×3 non - zero matrix U such that M2+MN2U is the zero matrix

(D) For a 3 ×3 matrix U, if M2+MN2U equals the zero matrix then U is the zero matrix

JEE Advanced · Mathematics · 18. Matrices

Let M and N be two $3 \times 3$ matrices such that MN = NM. Further, if $M \neq N^{2}$ and $M^{2} = N^{4}$ , then

A Determinant of $(M^{2} + M N^{2})$ is 0
B There is a $3 \times 3$ non - zero matrix $U$ such that $(M^{2} + M N^{2}) U$ is the zero matrix
C Determinant of $(M^{2} + M N^{2}) \geq 1$
D For a $3 \times 3$ matrix $U$ , if $(M^{2} + M N^{2}) U$ equals the zero matrix then $U$ is the zero matrix

Verified Solution

Answer & Solution

Correct Answer

(B) There is a $3 \times 3$ non - zero matrix $U$ such that $(M^{2} + M N^{2}) U$ is the zero matrix

Step-by-step Solution

Detailed explanation

M^{2} = N^{4} \Rightarrow M^{2} - N^{4} = 0 \Rightarrow (M - N^{2}) (M + N^{2}) = 0

(As M, N commute.)
Also,

M \neq N^{2}, D e t ((M - N^{2}) (M + N^{2})) = 0

M - N^{2}

is not null

\Rightarrow D e t (M + N^{2}) = 0

Also

D e t (M^{2} + M N^{2}) = (D e t M) (D e t (M + N^{2})) = 0

\Rightarrow

There exist non - null U such that

(M^{2} + M N^{2}) U = 0

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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