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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Matrices

NDA 2016 · Mathematics · Matrices

If A is a square matrix, then what is adj(A^ -1 )-(adj A)^ -1 equal...

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

Options

  1. A2∣A∣2|\mathrm{ A}|
  2. BNull matrix
  3. CUnit matrix
  4. DNone of the above

Correct answer

B. Null matrix

Step-by-step 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}adj⁡(A−1)=1(ad−bc)[acbd]T=1(ad−bc)[abcd]...(1)adj⁡A=[d−b−ca]∣adj⁡A∣=(ad−bc)(adj⁡A)−1=1∣adjA∣ [acbd]T=1ad−bc[abcd]....(2)\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), adj⁡(A−1)−(adj⁡A)−1=[0000]= null matrix. \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}
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