Q233 / 270 · NDA 2016 (Phase 1) · Mathematics

If A is a square matrix, then what is $\operatorname{adj}\left(\mathrm{A}^{-1}\right)-(\operatorname{adj} \mathrm{A})^{-1}$ equal to?

  • A
    $2|\mathrm{~A}|$
  • Correct answer
    B
    Null matrix
  • C
    Unit matrix
  • D
    None of the above

Solution

$\begin{aligned} & \text { Let }[A]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ & \begin{aligned} A^{-1} & =\frac{1}{|A|}(a d j . A) \\ & =\frac{1}{(a d-b c)}\left[\begin{array}{cc}d & -c \\ -b & a\end{array}\right]^T=\frac{1}{(a d-b c)}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\end{aligned}\end{aligned}$ $\begin{aligned} & \operatorname{adj}\left(A^{-1}\right)=\frac{1}{(a d-b c)}\left[\begin{array}{ll} a & c \\ b & d \end{array}\right]^T=\frac{1}{(a d-b c)}\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]...(1) \\ & \operatorname{adj} A=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \\ & |\operatorname{adj} A|=(a d-b c) \\ & (\operatorname{adj} A)^{-1}=\frac{1}{\mid a d j} A \left\lvert\,\left[\begin{array}{ll} a & c \\ b & d \end{array}\right]^T=\frac{1}{a d-b c}\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] .\right....(2) \end{aligned}$ Subtracting eqn. (1) and (2), $\begin{aligned} \operatorname{adj}\left(A^{-1}\right)-(\operatorname{adj} A)^{-1} & =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & =\text { null matrix. } \end{aligned}$

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

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