Mathematics · Matrices

JEE Main 2024 — 9 April, Shift 2 — Question 13

LetB=\left[ \begin{array}{*{35}{l}}1 & 3 \\1 & 5 \\\end{array} \right] and AA be a 2×22 \times 2 matrix such that AB−1=A−1\mathrm{AB}^{-1}=\mathrm{A}^{-1}. If BCB−1=A\mathrm{BCB}^{-1}=\mathrm{A} and C4+αC2+βI=O\mathrm{C}^{4}+\alpha \mathrm{C}^{2}+\beta \mathrm{I}=\mathrm{O}, then 2β−α2 \beta-\alpha is equal to :

  1. Option A:

    16

  2. Option B:

    2

  3. Option C:

    8

  4. Option D:

    10

    Correct

Answer: D

Step-by-step solution

BCB−1=A\mathrm{BCB}^{-1}=\mathrm{A}

⇒(BCB−1)(BCB−1)=A.A\Rightarrow\left(\mathrm{BCB}^{-1}\right)\left(\mathrm{BCB}^{-1}\right)=\mathrm{A} . \mathrm{A}

⇒BCICB−1=A2\Rightarrow \mathrm{BCICB}^{-1}=\mathrm{A}^{2}

⇒BC2 B−1=A2\Rightarrow \mathrm{BC}^{2} \mathrm{~B}^{-1}=\mathrm{A}^{2}

⇒B−1(BC2 B−1)B=B−1( A.A)B\Rightarrow \mathrm{B}^{-1}\left(\mathrm{BC}^{2} \mathrm{~B}^{-1}\right) \mathrm{B}=\mathrm{B}^{-1}(\mathrm{~A} . \mathrm{A}) \mathrm{B} From equation (1) C2=A−1⋅ A⋅ B\mathrm{C}^{2}=\mathrm{A}^{-1} \cdot \mathrm{~A} \cdot \mathrm{~B}

C2=B\mathrm{C}^{2}=\mathrm{B} Also AB−1=A−1\mathrm{AB}^{-1}=\mathrm{A}^{-1}

⇒AB−1⋅ A=A−1 A=I\Rightarrow \mathrm{AB}^{-1} \cdot \mathrm{~A}=\mathrm{A}^{-1} \mathrm{~A}=\mathrm{I}

⇒A−1(AB−1 A)=A−1I\Rightarrow \mathrm{A}^{-1}\left(\mathrm{AB}^{-1} \mathrm{~A}\right)=\mathrm{A}^{-1} \mathrm{I}

B−1 A=A−1\mathrm{B}^{-1} \mathrm{~A}=\mathrm{A}^{-1}

Now characteristics equation of C2\mathrm{C}^{2} is

=∣1−λ315−λ∣=0\left| \begin{matrix}1−\lambda & 3 \\1 & 5−\lambda \\\end{matrix} \right|=0 ⇒(1−λ)(5−1)−3=0⇒(λ2−6λ+5)−3=0\Rightarrow(1-\lambda)(5-1)-3=0 \Rightarrow\left(\lambda^{2}-6 \lambda+5\right)-3=0

⇒λ2−6λ+2=0\Rightarrow \lambda^{2}-6 \lambda+2=0

⇒β2−6 B+2I=0\Rightarrow \beta^{2}-6 \mathrm{~B}+2 \mathrm{I}=0

⇒C4−6C2+2I=0\Rightarrow C^{4}-6 C^{2}+2 I=0

α=−6\alpha=-6

β=2\beta=2

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

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Matrices
Topic
Inverse of a Matrix