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MCQMediumJEE 2024Inverse & Adjoint of a Matrix

JEE Mathematics 2024 Question with Solution

Let B=[1315]B = \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix} and AA be a 2×22 \times 2 matrix such that AB−1=A−1AB^{-1} = A^{-1}. If BCB−1=ABCB^{-1} = A and C4+αC2+βI=OC^4 + \alpha C^2 + \beta I = O, then 2β−α2\beta - \alpha is equal to:

  • A

    1616

  • B

    22

  • C

    88

  • D

    1010

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: B=[1315]B = \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}, AB−1=A−1AB^{-1} = A^{-1}, BCB−1=ABCB^{-1} = A, and C4+αC2+βI=OC^4 + \alpha C^2 + \beta I = O.

Find: 2β−α2\beta - \alpha.

From AB−1=A−1AB^{-1} = A^{-1}, post-multiply by BB:

A=A−1BA = A^{-1}B

Now pre-multiply by AA:

A2=BA^2 = B

Using BCB−1=ABCB^{-1} = A, square both sides:

(BCB−1)(BCB−1)=A2(BCB^{-1})(BCB^{-1}) = A^2 BC2B−1=A2=BBC^2B^{-1} = A^2 = B

Hence, multiplying by B−1B^{-1} on the left and by BB on the right,

C2=BC^2 = B

So C2C^2 satisfies the same characteristic equation as BB. Now,

∣B−λI∣=∣1−λ315−λ∣=0\left| B - \lambda I \right| = \begin{vmatrix} 1-\lambda & 3 \\ 1 & 5-\lambda \end{vmatrix} = 0 (1−λ)(5−λ)−3=0(1-\lambda)(5-\lambda) - 3 = 0 λ2−6λ+2=0\lambda^2 - 6\lambda + 2 = 0

Replacing λ\lambda by C2C^2, we get

C4−6C2+2I=OC^4 - 6C^2 + 2I = O

Comparing with

C4+αC2+βI=OC^4 + \alpha C^2 + \beta I = O

we obtain

α=−6,β=2\alpha = -6, \quad \beta = 2

Therefore,

2β−α=2×2−(−6)=102\beta - \alpha = 2 \times 2 - (-6) = 10

The correct option is D.

Using similarity and characteristic equation

Given: BCB−1=ABCB^{-1} = A shows that AA is similar to CC.

Find: the value of 2β−α2\beta - \alpha.

From the relation

AB−1=A−1AB^{-1} = A^{-1}

post-multiplying by BB gives

A=A−1BA = A^{-1}B

Then pre-multiplying by AA gives

A2=BA^2 = B

Now from similarity,

BCB−1=ABCB^{-1} = A

Squaring both sides,

BC2B−1=A2BC^2B^{-1} = A^2

Since A2=BA^2 = B,

BC2B−1=BBC^2B^{-1} = B

So,

C2=BC^2 = B

Thus the polynomial satisfied by BB will be satisfied by C2C^2. Compute the characteristic polynomial of BB:

det⁡(B−λI)=∣1−λ315−λ∣=(1−λ)(5−λ)−3=λ2−6λ+2\begin{aligned} \det(B-\lambda I) &= \begin{vmatrix} 1-\lambda & 3 \\ 1 & 5-\lambda \end{vmatrix} \\ &= (1-\lambda)(5-\lambda) - 3 \\ &= \lambda^2 - 6\lambda + 2 \end{aligned}

Hence,

B2−6B+2I=OB^2 - 6B + 2I = O

Since C2=BC^2 = B,

C4−6C2+2I=OC^4 - 6C^2 + 2I = O

Therefore,

α=−6,β=2\alpha = -6, \quad \beta = 2

and so

2β−α=4+6=102\beta - \alpha = 4 + 6 = 10

Therefore, the correct option is D.

Common mistakes

  • From AB−1=A−1AB^{-1} = A^{-1}, concluding A=BA = B is incorrect. The correct manipulation is to multiply by BB and then by AA to get A2=BA^2 = B.

  • While squaring BCB−1=ABCB^{-1} = A, forgetting that B−1B=IB^{-1}B = I can lead to an incorrect expression. The correct result is BC2B−1=A2BC^2B^{-1} = A^2.

  • After obtaining the characteristic polynomial of BB, substituting directly for CC instead of C2C^2 is wrong. Since C2=BC^2 = B, the polynomial must be written in terms of C2C^2.

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