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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
Given that $A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & 1 \\ -3 & 2\end{array}\right]$, matrix $A$ is :
  • A
    $7\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
  • B
    $\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
  • C
    $\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
  • D
    $\frac{1}{49}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
✓

Answer

We have, $A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2 & 1 \\ -3 & 2\end{array}\right]$
$\because\left(A^{-1}\right)^{-1}=A$
Now, $\left(A^{-1}\right)^{-1}=\frac{\operatorname{adj}\left(A^{-1}\right)}{\left|A^{-1}\right|}=\frac{\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]}{\frac{1}{7}}=\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
$
\Rightarrow \quad A=\left[\begin{array}{cc}
2 & -1 \\
3 & 2\end{array}\right]
$

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