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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
If $A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&1\end{array}} \right],$ then $adj\;\left( {3{A^2} + 12A} \right) = $ . . . .
  • A
    $\left[ {\begin{array}{*{20}{c}}{72}&{ - 63}\\{ - 84}&{51}\end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}{72}&{ - 84}\\{ - 63}&{51}\end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}{51}&{63}\\{84}&{72}\end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}{51}&{84}\\{63}&{72}\end{array}} \right]$
✓

Answer

Correct option: C.
$\left[ {\begin{array}{*{20}{c}}{51}&{63}\\{84}&{72}\end{array}} \right]$
c
We have $A = \left[ {\begin{array}{*{20}{c}}
2&{ - 3}\\
{ - 4}&1
\end{array}} \right]$

$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{16}&{ - 9}\\
{ - 12}&{13}
\end{array}} \right]$

$ \Rightarrow 3{A^2} = \left[ {\begin{array}{*{20}{c}}
{48}&{ - 27}\\
{ - 36}&{39}
\end{array}} \right]$

Also $12A = \left[ {\begin{array}{*{20}{c}}
{24}&{ - 36}\\
{ - 48}&{12}
\end{array}} \right]$

$\therefore 3{A^2} + 12A = \left[ {\begin{array}{*{20}{c}}
{48}&{ - 27}\\
{ - 36}&{39}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{24}&{ - 36}\\
{ - 48}&{12}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{72}&{ - 63}\\
{ - 84}&{51}
\end{array}} \right]$

adj $\left( {3{A^2} + 12A} \right) = \left[ {\begin{array}{*{20}{c}}
{51}&{63}\\
{84}&{72}
\end{array}} \right]$

 

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