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Matrices and Determinants question

2022 · 28 Jul · Shift 2 · Q23
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Matrices and Determinants question

2022 · 28 Jul · Shift 2 · Q23

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A\mathrm{A}A and B\mathrm{B}B be any two 3×33 \times 33×3 symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?
  1. A
    A4−B4\mathrm{A}^{4}-\mathrm{B}^{4}A4−B4 is a smmetric matrix
  2. B
    AB−BA\mathrm{AB}-\mathrm{BA}AB−BA is a symmetric matrix
  3. C
    B5−A5\mathrm{B}^{5}-\mathrm{A}^{5}B5−A5 is a skew-symmetric matrix
  4. D
    AB+BA\mathrm{AB}+\mathrm{BA}AB+BA is a skew-symmetric matrix
View written solutionFree

Correct answer: C

  1. Given properties

For a symmetric matrix AAA and a skew-symmetric matrix BBB:

  • AT=AA^T = AAT=A
  • BT=−BB^T = -BBT=−B

We use:

  • (X+Y)T=XT+YT(X+Y)^T = X^T + Y^T(X+Y)T=XT+YT
  • (XY)T=YTXT(XY)^T = Y^T X^T(XY)T=YTXT
  • (Xn)T=(XT)n(X^n)^T = (X^T)^n(Xn)T=(XT)n

Also,

  • MMM is symmetric if MT=MM^T = MMT=M
  • MMM is skew-symmetric if MT=−MM^T = -MMT=−M
  1. Check Option A: A4−B4A^4 - B^4A4−B4 is symmetric

First,

(A4)T=(AT)4=A4(A^4)^T = (A^T)^4 = A^4(A4)T=(AT)4=A4

so A4A^4A4 is symmetric.

Now,

(B4)T=(BT)4=(−B)4=B4(B^4)^T = (B^T)^4 = (-B)^4 = B^4(B4)T=(BT)4=(−B)4=B4

so B4B^4B4 is also symmetric.

Difference of two symmetric matrices is symmetric, hence

(A4−B4)T=A4−B4(A^4 - B^4)^T = A^4 - B^4(A4−B4)T=A4−B4

So Option A is true.

  1. Check Option B: AB−BAAB - BAAB−BA is symmetric

Take transpose:

(AB−BA)T=(AB)T−(BA)T=BTAT−ATBT(AB-BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T(AB−BA)T=(AB)T−(BA)T=BTAT−ATBT

Using AT=AA^T=AAT=A and BT=−BB^T=-BBT=−B,

(AB−BA)T=(−B)A−A(−B)=−BA+AB=AB−BA(AB-BA)^T = (-B)A - A(-B) = -BA + AB = AB - BA(AB−BA)T=(−B)A−A(−B)=−BA+AB=AB−BA

Thus,

(AB−BA)T=AB−BA(AB-BA)^T = AB-BA(AB−BA)T=AB−BA

So AB−BAAB-BAAB−BA is symmetric.

Hence Option B is true.

  1. Check Option C: B5−A5B^5 - A^5B5−A5 is skew-symmetric

First,

(B5)T=(BT)5=(−B)5=−B5(B^5)^T = (B^T)^5 = (-B)^5 = -B^5(B5)T=(BT)5=(−B)5=−B5

so B5B^5B5 is skew-symmetric.

Also,

(A5)T=(AT)5=A5(A^5)^T = (A^T)^5 = A^5(A5)T=(AT)5=A5

so A5A^5A5 is symmetric.

Now consider

M=B5−A5M = B^5 - A^5M=B5−A5

Then

MT=(B5−A5)T=(B5)T−(A5)T=−B5−A5M^T = (B^5 - A^5)^T = (B^5)^T - (A^5)^T = -B^5 - A^5MT=(B5−A5)T=(B5)T−(A5)T=−B5−A5

For MMM to be skew-symmetric, we need

MT=−M=−(B5−A5)=−B5+A5M^T = -M = -(B^5 - A^5) = -B^5 + A^5MT=−M=−(B5−A5)=−B5+A5

But in general,

−B5−A5≠−B5+A5-B^5 - A^5 \neq -B^5 + A^5−B5−A5=−B5+A5

unless A5=0A^5=0A5=0.

So in general B5−A5B^5-A^5B5−A5 is not skew-symmetric.

Hence Option C is NOT true.

  1. Check Option D: AB+BAAB+BAAB+BA is skew-symmetric

Take transpose:

(AB+BA)T=(AB)T+(BA)T=BTAT+ATBT(AB+BA)^T = (AB)^T + (BA)^T = B^T A^T + A^T B^T(AB+BA)T=(AB)T+(BA)T=BTAT+ATBT

Using AT=AA^T=AAT=A, BT=−BB^T=-BBT=−B,

(AB+BA)T=(−B)A+A(−B)=−BA−AB=−(AB+BA)(AB+BA)^T = (-B)A + A(-B) = -BA - AB = -(AB+BA)(AB+BA)T=(−B)A+A(−B)=−BA−AB=−(AB+BA)

Thus AB+BAAB+BAAB+BA is skew-symmetric.

So Option D is true.

  1. Conclusion

The only statement which is NOT true is:

C\boxed{\text{C}}C​
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