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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
If $A = \left[ {\begin{array}{*{20}{c}}{ - 1}&2\\2&{ - 1}\end{array}} \right]$ and $B = \left[ \begin{array}{l}3\\1\end{array} \right],AX = B$, then $X = $
  • A
    $[5 7]$
  • ✓
    $\frac{1}{3}\left[ \begin{array}{l}5\\7\end{array} \right]$
  • C
    $\frac{1}{3}[5\,\,7]$
  • D
    $\left[ \begin{array}{l}5\\7\end{array} \right]$
✓

Answer

Correct option: B.
$\frac{1}{3}\left[ \begin{array}{l}5\\7\end{array} \right]$
b
(b) $A = \left[ {\,\begin{array}{*{20}{c}}{ - 1}&2\\2&{ - 1}\end{array}\,} \right]\,\,,\,\,\,B = \left[ \begin{array}{l}3\\1\end{array} \right]$

$AX = B\, \Rightarrow X = {A^{ - 1}}B$

${A^{ - 1}} = \frac{{adj\,A}}{{|A|}}$

${A^{ - 1}} = \frac{{ - 1}}{3}\left[ {\,\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 2}&{ - 1}\end{array}\,} \right]$$ = \frac{1}{3}\left[ {\,\begin{array}{*{20}{c}}1&2\\2&1\end{array}\,} \right]$

and $X = {A^{ - 1}}B$ = $\left[ {\,\begin{array}{*{20}{c}}{\frac{1}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}\end{array}\,} \right]\,\left[ \begin{array}{l}3\\1\end{array} \right]$; $X = \frac{1}{3}\left[ \begin{array}{l}5\\7\end{array} \right]$.

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