JEE Advanced 2014 · Paper 1 · Mathematics Q41

MathematicsMultiple correct+3Question 41 of 60

Let $M$ and $N$ be two $3\times3$ matrices such that $MN=NM$ . Further, if $M\ne N^2$ and $M^2=N^4$ , then

determinant of

(M^2+MN^2)

is 0

there is a

3\times3

non-zero matrix

U

such that

(M^2+MN^2)U

is the zero matrix

determinant of

(M^2+MN^2)\ge1

for a

3\times3

matrix

U

, if

(M^2+MN^2)U

equals the zero matrix then

U

is the zero matrix

Answer and worked solution

Official answer: A, B

Factorise. Since $M$ and $N$ commute, so do $M$ and $N^2$ :
$M^2-N^4=(M-N^2)(M+N^2)=0.$

$M+N^2$ is singular. If it were invertible, multiplying on the right by its inverse would give $M=N^2$ , which is ruled out. So $\det(M+N^2)=0$ .

The matrix in question. $M^2+MN^2=M(M+N^2)$ , so $\det(M^2+MN^2)=\det M\cdot0=0$ .

(A) True.
(B) A singular matrix has a non-trivial null space, so a non-zero $U$ with $(M^2+MN^2)U=0$ exists (take its columns in the null space). True.
(C) The determinant is 0. False.
(D) Contradicts (B). False.

(A), (B)

Topics:Matrix algebra Determinants and properties

JEE Advanced 2014 · Paper 1, Mathematics question 41