JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Matrices

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

(A)determinant of $(M^{2} + M N^{2})$ is 0✓ Correct
(B)there is a $3 \times 3$ non-zero matrix $U$ such that $(M^{2} + M N^{2}) U$ is the zero matrix✓ Correct
(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

Answer: (A), (B)

Solution

Since

M N = N M

, we have

M N^{2} = N^{2} M

. Now

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

(using

M N^{2} = N^{2} M

). Given

M^{2} = N^{4}

, we get

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

. Since

M \neq = N^{2}

, the matrix

M - N^{2}

is non-zero, which implies

∣ M + N^{2} ∣ = 0

. Therefore

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

, so option (A) is correct. Since

∣ M + N^{2} ∣ = 0

, there exists a non-zero matrix

U

such that

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

, hence

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

, so option (B) is correct.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options