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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Matrices

NDA 2021 · Mathematics · Matrices

The inverse of a matrix A is given by [ array cc -2 & 1 (3)/(2) &...

The inverse of a matrix AA is given by \left[−2132−12\begin{array}{cc} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{array}\right] What is AA equal to?

Options

  1. A\left[1234\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]
  2. B\left[1−2−34\begin{array}{cc}1 & -2 \\ -3 & 4\end{array}\right]
  3. C\left[123−4\begin{array}{cc}1 & 2 \\ 3 & -4\end{array}\right]
  4. D\left[−1234\begin{array}{cc}-1 & 2 \\ 3 & 4\end{array}\right]

Correct answer

A. \left[1234\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]

Step-by-step solution

A=(A−1)−1=1(−2×−12−1×32)[−12−1−32−2]=−2[−12−1−32−2]=[1234]\begin{aligned} & A=\left(A^{-1}\right)^{-1}=\frac{1}{\left(-2 \times-\frac{1}{2}-1 \times \frac{3}{2}\right)}\left[\begin{array}{ll}-\frac{1}{2} & -1 \\ -\frac{3}{2} & -2\end{array}\right] \\ & =-2\left[\begin{array}{ll}-\frac{1}{2} & -1 \\ -\frac{3}{2} & -2\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\end{aligned}
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