दिया है,  $A = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&{ – 1}\end{array}} \right],\,\,B = \left[ {\begin{array}{*{20}{c}}a&1\\b&{ – 1}\end{array}} \right]$   

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

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

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

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

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

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

अब,   ${(A + B)^2} = {A^2} + {B^2}$ 

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

  तुलना करने पर, $a – 1 = 0 \Rightarrow a = 1$ तथा $b = 4$.