Q69 / 155 · WBJEE 2023 · Mathematics

Let \(A\) and \(B\) are orthogonal matrices and \(\operatorname{det} A+\operatorname{det} B=0\). Then

  • Correct answer
    A
    \(A+B\) is singular
  • B
    \(A+B\) is non-singular
  • C
    \(A+B\) is orthogonal
  • D
    \(A+B\) is skew symmetric

Solution

Hint: \(A A^{\top}=I\) and \(B B^{\top}=I\) \(\Rightarrow \operatorname{det}\left(A A^{\top}\right)=1\) and \(\operatorname{det}\left(B B^{\top}\right)=1\) \(\operatorname{det}(A)=-\operatorname{det}(B)\) (Given) \( \Rightarrow(\operatorname{det}(A))^2=1 \Rightarrow(\operatorname{det}(B))^2=1\) \(\therefore \operatorname{det}(A+B)=\operatorname{det}\left(A\left(B^{\top}+A^{\top}\right) B\right)\) \(=-\operatorname{det}\left(B^{\top}+A^{\top}\right) \quad[\because(\operatorname{det}(A))(\operatorname{det}(B))=-1]\) as \(\operatorname{det}(A)=-\operatorname{det}(B)\) \(=-\operatorname{det}(B+A)^{\top}\) \(=-\operatorname{det}(A+B)^{\top}\) \(=-\operatorname{det}(A+B)\) \(\Rightarrow 2 \operatorname{det}(A+B)=0\) \(\Rightarrow \operatorname{det}(A+B)=0\) \(\therefore(A+B)\) is singular

Hence, the correct answer is option: 'A'.

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