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

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determinant of (M²+ MN²) is 0

There is a 3 × 3 non-zero matrix U such that $(M^{2} + {MN}^{2}) U$ is zero matrix

determinant of $(M^{2} + {MN}^{2}) \geq 1$

for a 3 × 3 matrix u, if $(M^{2} + {MN}^{2}) U$ equal to the zero matrix, then U is the zero matrix

answer is A.

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Detailed Solution

(i) If A and B are two non-zero matrices and AB=BA, then (A–B)(A+B)=A²–B^2.
(ii) The determinant of the product of the matrices is equal to product of their individual determinants i.e $| AB | = | A | | B |$
Given, $M^{2} = N^{4} \Rightarrow M^{2} - N^{4} = 0$ $\Rightarrow (M - N^{2}) (M + N^{2}) = 0 (as MN = NM)$
Also, $M \neq N^{2} \Rightarrow M + N^{2} = 0 \Rightarrow Det (M + N^{2}) = 0$
Also, $Det (M^{2} + {MN}^{2}) = (Det M) (Det (M + N^{2}))$
$= (Det M) (0) = 0 As, Det (M^{2} + {MN}^{2}) = 0$