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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Matrices

NDA 2012 · Mathematics · Matrices

If A=[ array ll 1 & 2 1 & 1 array ] and B=[ array cc 0 & -1 1 & 2...

If A=\left[1211\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right] andB=\left[0−112\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right],thenwhatis, then what isB^{-1}\mathrm{A}^{-1}$ equal to?

Options

  1. A\left[1−3−12\begin{array}{cc}1 & -3 \\ -1 & 2\end{array}\right]
  2. B\left[−131−2\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]
  3. C\left[−13−1−2\begin{array}{cc}-1 & 3 \\ -1 & -2\end{array}\right]
  4. D\left[−1−31−2\begin{array}{cc}-1 & -3 \\ 1 & -2\end{array}\right]

Correct answer

B. \left[−131−2\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]

Step-by-step solution

∣A∣=−1,∣B∣=1A−1=1∣A∣Adj⁡A=1−1[1−1−21]T=[−121−1]\begin{aligned} |\mathrm{A}| & =-1,|\mathrm{ B}|=1 \\ \mathrm{ A}^{-1} & =\frac{1}{|\mathrm{ A}|} \operatorname{Adj} \mathrm{A}=\frac{1}{-1}\left[\begin{array}{cc} 1 & -1 \\ -2 & 1 \end{array}\right]^{\mathrm{T}} \\ & =\left[\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right] \end{aligned}B−1=1∣B∣Adj=11[2−110]T=[21−10]B−1A−1=[21−10][−121−1]=[−131−2].\begin{aligned} & \mathrm{B}^{-1}=\frac{1}{|\mathrm{ B}|} \mathrm{Adj}=\frac{1}{1}\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc} 2 & 1 \\ -1 & 0 \end{array}\right] \\ & \mathrm{B}^{-1} \mathrm{ A}^{-1}=\left[\begin{array}{cc} 2 & 1 \\ -1 & 0 \end{array}\right]\left[\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{cc} -1 & 3 \\ 1 & -2 \end{array}\right]. \end{aligned}
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