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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 $A=\left[\begin{array}{cc}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right],$ then find the matrix $X$, such that $2 \mathrm{A}+3 \mathrm{X}=5 \mathrm{B}$.
  • A
    $\left[ {\begin{array}{*{20}{c}}
      {  2}&{\frac{{ - 10}}{3}} \\ 
      4&{\frac{{14}}{3}} \\ 
      {\frac{{ - 31}}{3}}&{\frac{{  7}}{3}} 
    \end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}
      { 2}&{\frac{{  10}}{3}} \\ 
      4&{\frac{{14}}{3}} \\ 
      {\frac{{ - 31}}{3}}&{\frac{{  7}}{3}} 
    \end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}
      { - 2}&{\frac{{  10}}{3}} \\ 
      4&{\frac{{-14}}{3}} \\ 
      {\frac{{ - 31}}{3}}&{\frac{{  7}}{3}} 
    \end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}
      { - 2}&{\frac{{ - 10}}{3}} \\ 
      4&{\frac{{14}}{3}} \\ 
      {\frac{{ - 31}}{3}}&{\frac{{ - 7}}{3}} 
    \end{array}} \right]$
✓

Answer

Correct option: D.
$\left[ {\begin{array}{*{20}{c}}
  { - 2}&{\frac{{ - 10}}{3}} \\ 
  4&{\frac{{14}}{3}} \\ 
  {\frac{{ - 31}}{3}}&{\frac{{ - 7}}{3}} 
\end{array}} \right]$
d
We have $2A+3X=5 B$

or         $2A+3X-2A=5B-2A$

or         $2 A-2A+3X=5B-2A$          $($ Matrix addition is commutative $)$

or         $O+3 X=5B-2 A$        $(-2A$ is the additive inverse of $2A)$

or         $3 \mathrm{X}=5 \mathrm{B}-2 \mathrm{A}$                  ( $O$ is the additive identity)

or         $X=\frac{1}{3}(5 B-2 A)$

or         ${\text{X}} = $ $\frac{1}{3}\left( {5\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\ 
  4&2 \\ 
  { - 5}&1 
\end{array}} \right] - 2\left[ {\begin{array}{*{20}{l}}
  8&0 \\ 
  4&{ - 2} \\ 
  3&6 
\end{array}} \right]} \right)$ $ = \frac{1}{3}\left( {\left[ {\begin{array}{*{20}{c}}
  {10}&{ - 10} \\ 
  {20}&{10} \\ 
  { - 25}&5 
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  { - 16}&0 \\ 
  { - 8}&4 \\ 
  { - 6}&{ - 12} 
\end{array}} \right]} \right)$

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
  {10 - 16}&{ - 10 + 0} \\ 
  {20 - 8}&{10 + 4} \\ 
  { - 25 - 6}&{5 - 12} 
\end{array}} \right]$ 

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
  { - 6}&{ - 10} \\ 
  {12}&{14} \\ 
  { - 31}&{ - 7} 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{\frac{{ - 10}}{3}} \\ 
  4&{\frac{{14}}{3}} \\ 
  {\frac{{ - 31}}{3}}&{\frac{{ - 7}}{3}} 
\end{array}} \right]$

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