JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 15)

Let M and N be two 3 $$\times$$ 3 matrices such that MN = NM. Further, if M $$\ne$$ N2 and M2 = N4, then
determinant of (M2 + MN2) is 0
there is a 3 $$\times$$ 3 non-zero matrix U such that (M2 + MN2) U is zero matrix
determinant of (M2 + MN2) $$\ge$$ 1
for a 3 $$\times$$ 3 matrix U, if (M2 + MN2) U equals the zero matrix, then U is the zero matrix

Explanation

Given MN = NM. Therefore, a2 $$-$$ b2 = (a + b) (a $$-$$ b) of algebra of numbers is applicable.

Now, M2 = N4 $$\Rightarrow$$ M2 $$-$$ N4 = 0 (Null matrix)

$$\Rightarrow$$ (M + N2) (M $$-$$ N2) = 0

Since M $$\ne$$ N2 (given), the possibilities are

(M + N2) = 0 and M $$-$$ N2 $$\ne$$ 0 (1)

or (M + N2) $$\ne$$ 0 and M + N2 $$\ne$$ 0 (2)

Now, we know if A and B are non-null square matrix nd AB = 0, then A and B both are singular, that is, | A | = 0 and | B | = 0 and AB = 0.

Note : For example, let A be non-singular. Therefore, B = In(2) = A$$-$$1 AB = 0 (since AB = 0 is assumed).

Hence, B is singular, which is a contradiction and A has to be singular. Similarly, B also has to be singular.

Therefore, from Eqs. (1) and (2), we conclude the only possibility is | M + N2 | = 0.

Now checking options :

(A) | M2 + MN2 | = | M || M + N2 | = 0

Hence, option (A) is correct.

(B) (M2 + MN2) U = 0

Since, M2 + MN2 is singular. Therefore, U has infinitely many possible values (non-trivial solutions). Hence, option (B) is true.

(C) False since | M2 + MN2 | = 0.

(D) Since | M2 + MN2 | = 0, U is not a necessarily a zero matrix.

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