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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 $A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&{ - 1}\end{array}} \right],\,\,B = \left[ {\begin{array}{*{20}{c}}a&1\\b&{ - 1}\end{array}} \right]$ and ${(A + B)^2} = {A^2} + {B^2}$, then the value of $a$ and $b$ are
  • A
    $a = 4,b = 1$
  • ✓
    $a = 1,b = 4$
  • C
    $a = 0,b = 4$
  • D
    $a = 2,b = 4$
✓

Answer

Correct option: B.
$a = 1,b = 4$
b
(b) Given, $A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&{ - 1}\end{array}} \right],\,\,B = \left[ {\begin{array}{*{20}{c}}a&1\\b&{ - 1}\end{array}} \right]$

==> $A + B = \left[ {\begin{array}{*{20}{c}}{1 + a}&0\\{2 + b}&{ - 2}\end{array}} \right]$

${A^2} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&{ - 1}\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&{ - 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]$

${B^2} = \left[ {\begin{array}{*{20}{c}}a&1\\b&{ - 1}\end{array}} \right]\,\,\,\left[ {\begin{array}{*{20}{c}}a&1\\b&{ - 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2} + b}&{a - 1}\\{ab - b}&{b + 1}\end{array}} \right]$

Also, ${A^2} + {B^2} = \left[ {\begin{array}{*{20}{c}}{{a^2} + b - 1}&{a - 1}\\{ab - b}&b\end{array}} \right]$

==> ${(A + B)^2} = \left[ {\begin{array}{*{20}{c}}{1 + a}&0\\{2 + b}&{ - 2}\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{1 + a}&0\\{2 + b}&{ - 2}\end{array}} \right]$

Also, ${(A + B)^2} = \left[ {\begin{array}{*{20}{c}}{{{(1 + a)}^2}}&{\,\,\,\,0}\\{(2 + b)(1 + a) - 2(2 + b)}&{\,\,\,\,4}\end{array}} \right]$

Also, ${(A + B)^2} = {A^2} + {B^2}$

==> $\left[ {\begin{array}{*{20}{c}}{{{(1 + a)}^2}}&0\\{\,(2 + b)(a - 1)}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2} + b - 1}&{a - 1}\\{ab - b}&b\end{array}} \right]$

By equating, $a - 1 = 0 \Rightarrow a = 1$ and $b = 4$.

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