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Matrices and Determinants question

2013 · Shift 1 · Q33
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  5. /2013 · Shift 1 · Q33

Matrices and Determinants question

2013 · Shift 1 · Q33

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −1
For 3 × 3 matrices M and N, which of the following statement(s) is(are) NOT correct?
  1. A
    NTMN is symmetric or skew symmetric, according as M is symmetric or skew symmetric.
  2. B
    MN – NM is skew symmetric for all symmetric matrices M and N.
  3. C
    MN is symmetric for all symmetric matrices M and N.
  4. D
    (adj M)·(adj N) = adj(MN) for all invertible matrices M and N.
View written solutionFree

Correct answer: C, D

We check each statement one by one.

1. Option A

Statement:

NTMN is symmetric or skew-symmetric, according as M is symmetric or skew-symmetric.N^T M N \text{ is symmetric or skew-symmetric, according as } M \text{ is symmetric or skew-symmetric.}NTMN is symmetric or skew-symmetric, according as M is symmetric or skew-symmetric.

We use transpose:

(NTMN)T=NTMTN\left(N^T M N\right)^T = N^T M^T N(NTMN)T=NTMTN

Case 1: MMM is symmetric

Then

MT=MM^T = MMT=M

so

(NTMN)T=NTMN\left(N^T M N\right)^T = N^T M N(NTMN)T=NTMN

Hence NTMNN^T M NNTMN is symmetric.

Case 2: MMM is skew-symmetric

Then

MT=−MM^T = -MMT=−M

so

(NTMN)T=NT(−M)N=−NTMN\left(N^T M N\right)^T = N^T (-M) N = -N^T M N(NTMN)T=NT(−M)N=−NTMN

Hence NTMNN^T M NNTMN is skew-symmetric.

So A is correct.

2. Option B

Statement:

MN−NM is skew-symmetric for all symmetric matrices M,N.MN - NM \text{ is skew-symmetric for all symmetric matrices } M,N.MN−NM is skew-symmetric for all symmetric matrices M,N.

Since M,NM,NM,N are symmetric,

MT=M,NT=NM^T = M, \qquad N^T = NMT=M,NT=N

Now,

(MN−NM)T=(MN)T−(NM)T=NTMT−MTNT=NM−MN\left(MN - NM\right)^T = (MN)^T - (NM)^T = N^T M^T - M^T N^T = NM - MN(MN−NM)T=(MN)T−(NM)T=NTMT−MTNT=NM−MN

Thus,

(MN−NM)T=−(MN−NM)\left(MN - NM\right)^T = -(MN - NM)(MN−NM)T=−(MN−NM)

Hence MN−NMMN-NMMN−NM is skew-symmetric.

So B is correct.

3. Option C

Statement:

MN is symmetric for all symmetric matrices M,N.MN \text{ is symmetric for all symmetric matrices } M,N.MN is symmetric for all symmetric matrices M,N.

For MNMNMN to be symmetric, we need

(MN)T=MN(MN)^T = MN(MN)T=MN

But

(MN)T=NTMT=NM(MN)^T = N^T M^T = NM(MN)T=NTMT=NM

So MNMNMN is symmetric iff

NM=MNNM = MNNM=MN

That is, iff MMM and NNN commute. Two symmetric matrices need not commute in general.

Counterexample

Take

N=(110110000)\qquad N = \begin{pmatrix}1&1&0\\1&1&0\\0&0&0\end{pmatrix}N=​110​110​000​​

Both are symmetric.

Now,

MN=(110000000)MN = \begin{pmatrix}1&1&0\\0&0&0\\0&0&0\end{pmatrix}MN=​100​100​000​​

and

(MN)T=(100100000)≠MN(MN)^T = \begin{pmatrix}1&0&0\\1&0&0\\0&0&0\end{pmatrix} \neq MN(MN)T=​110​000​000​​=MN

So MNMNMN is not symmetric.

Therefore C is NOT correct.

4. Option D

Statement:

(adj⁡M)(adj⁡N)=adj⁡(MN)for all invertible matrices M,N.(\operatorname{adj} M)(\operatorname{adj} N) = \operatorname{adj}(MN) \quad \text{for all invertible matrices } M,N.(adjM)(adjN)=adj(MN)for all invertible matrices M,N.

For invertible matrices,

adj⁡(A)=(det⁡A)A−1\operatorname{adj}(A) = (\det A)A^{-1}adj(A)=(detA)A−1

Thus,

= (\det M)M^{-1}(\det N)N^{-1} = (\det M)(\det N) M^{-1}N^{-1}$$ But $$\operatorname{adj}(MN) = \det(MN)(MN)^{-1} = (\det M)(\det N)(MN)^{-1}$$ and $$(MN)^{-1} = N^{-1}M^{-1}$$ So $$\operatorname{adj}(MN) = (\det M)(\det N)N^{-1}M^{-1}$$ In general, $$M^{-1}N^{-1} \neq N^{-1}M^{-1}$$ unless $M$ and $N$ commute. Hence, in general, $$(\operatorname{adj} M)(\operatorname{adj} N) \neq \operatorname{adj}(MN)$$ In fact, the correct identity is $$\operatorname{adj}(MN) = (\operatorname{adj} N)(\operatorname{adj} M)$$ So **D is NOT correct**. --- ## Final conclusion The statements which are **NOT correct** are: $$\boxed{C, D}$$ --- ## Comparison with stored correct answer Stored correct answer: $C, D$ Our derived answer matches the stored answer.
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