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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
If $A=\left[\begin{array}{ccc}1 & -2 & 4 \\ 2 & -1 & 3 \\ 4 & 2 & 0\end{array}\right]$ is the adjoint of a square matrix $B$, then $B^{-1}$ is equal to
  • A
    $\pm A$
  • B
    $\pm \sqrt{2} A$
  • C
    $\pm \frac{1}{\sqrt{2}} B$
  • ✓
    $\pm \frac{1}{\sqrt{2}} A$
✓

Answer

Correct option: D.
$\pm \frac{1}{\sqrt{2}} A$
$A|=\left|\begin{array}{ccc}1 & -2 & 4 \\ 2 & -1 & 3 \\ 4 & 2 & 0\end{array}\right| $
$=1(0-6)+2(0-12)+4(4+4)=2$
Given $,  A=\operatorname{adj} B$
$\Rightarrow|A|=|\operatorname{adj} B| $
$\Rightarrow|\operatorname{adj} B|=2$
$\Rightarrow|B|^2=2 \quad\left[\because|\operatorname{adj} B|=|B|^{3-1} \text {, where } B \text { is } 3 \times 3 \text { matrix }\right]$
$\Rightarrow|B|= \pm \sqrt{2}$
$\therefore B^{-1}= \pm \frac{1}{\sqrt{2}} A\left[\because B^{-1}=\frac{1}{|B|}(\operatorname{adj} B)\right]$

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