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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
For $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then $14 A^{-1}$ is given by
  • A
    $14\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
  • B
    $\left[\begin{array}{cc}4 & -2 \\ 2 & 6\end{array}\right]$
  • C
    $2\left[\begin{array}{ll}2 & -1 \\ 1 & -3\end{array}\right]$
  • D
    $2\left[\begin{array}{cc}-3 & -1 \\ 1 & -2\end{array}\right]$
✓

Answer

We have, $|A|=6+1=7$
Also, $\operatorname{adj} A=\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
Now, $A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$
$
\therefore 14 A^{-1}=\left[\begin{array}{cc}
4 & -2 \\
2 & 6
\end{array}\right]
$

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