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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
$a = 4,b = 1$
$a = 1,b = 4$
$a = 0,b = 4$
$a = 2,b = 4$
Solution
(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 $b = 4$.
