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Question Number 164134 by Avijit007 last updated on 14/Jan/22

	Answered by Rasheed.Sindhi last updated on 14/Jan/22
	$A= [(a_(11) ,a_(12) ),(a_(21) ,a_(22) ) ],B= [(b_(11) ,b_(12) ),(b_(21) ,b_(22) ) ](say) •2A+B^t = [((2a_(11) ),(2a_(12) )),((2a_(21) ),(2a_(22) )) ]+ [(b_(11) ,b_(21) ),(b_(12) ,b_(22) ) ]= [(2,5),((10),2) ] [((2a_(11) +b_(11) ),(2a_(12) +b_(21) )),((2a_(21) +b_(12) ),(2a_(22) +b_(22) )) ]= [(2,5),((10),2) ] •2B+A^t = [((2b_(11) ),(2b_(12) )),((2b_(21) ),(2b_(22) )) ]+ [(a_(11) ,a_(21) ),(a_(12) ,a_(22) ) ]= [(1,8),(4,1) ] [((2b_(11) +a_(11) ),(2b_(12) +a_(21) )),((2b_(21) +a_(12) ),(2b_(22) +a_(22) )) ]= [(1,8),(4,1) ] { ((2a_(11) +b_(11) =2⇒2a_(11) +b_(11) =2)),((2b_(11) +a_(11) =1⇒2a_(11) +4b_(11) =2)) :} ⇒3b_(11) =0⇒b_(11) =0⇒a_(11) =1 { ((2a_(22) +b_(22) =2⇒2a_(22) +b_(22) =2)),((2b_(22) +a_(22) =1⇒2a_(22) +4b_(22) =2)) :} ⇒3b_(22) =0⇒b_(22) =0⇒a_(22) =1 { ((2a_(12) +b_(21) =5⇒2a_(12) +b_(21) =5)),((2b_(21) +a_(12) =4⇒2a_(12) +4b_(21) =8)) :} 3b_(21) =3⇒b_(21) =1⇒a_(12) =2 { ((2a_(21) +b_(12) =10⇒2a_(21) +b_(12) =10)),((2b_(12) +a_(21) =8⇒2a_(21) +4b_(12) =16)) :} 3b_(12) =6⇒b_(12) =2⇒a_(21) =4 A= [(a_(11) ,a_(12) ),(a_(21) ,a_(22) ) ]= [(1,2),(4,1) ] B= [(b_(11) ,b_(12) ),(b_(21) ,b_(22) ) ]= [(0,2),(1,0) ]$
	$A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}], B = [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] (s a y)$ $∙ 2 A + B^{t} = [\begin{matrix} 2 a_{11} & 2 a_{12} \\ 2 a_{21} & 2 a_{22} \end{matrix}] + [\begin{matrix} b_{11} & b_{21} \\ b_{12} & b_{22} \end{matrix}] = [\begin{matrix} 2 & 5 \\ 10 & 2 \end{matrix}]$ $[\begin{matrix} 2 a_{11} + b_{11} & 2 a_{12} + b_{21} \\ 2 a_{21} + b_{12} & 2 a_{22} + b_{22} \end{matrix}] = [\begin{matrix} 2 & 5 \\ 10 & 2 \end{matrix}]$ $∙ 2 B + A^{t} = [\begin{matrix} 2 b_{11} & 2 b_{12} \\ 2 b_{21} & 2 b_{22} \end{matrix}] + [\begin{matrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{matrix}] = [\begin{matrix} 1 & 8 \\ 4 & 1 \end{matrix}]$ $[\begin{matrix} 2 b_{11} + a_{11} & 2 b_{12} + a_{21} \\ 2 b_{21} + a_{12} & 2 b_{22} + a_{22} \end{matrix}] = [\begin{matrix} 1 & 8 \\ 4 & 1 \end{matrix}]$ ${\begin{cases} 2 a_{11} + b_{11} = 2 \Rightarrow 2 a_{11} + b_{11} = 2 \\ 2 b_{11} + a_{11} = 1 \Rightarrow 2 a_{11} + 4 b_{11} = 2 \end{cases}$ $\Rightarrow 3 b_{11} = 0 \Rightarrow b_{11} = 0 \Rightarrow a_{11} = 1$ ${\begin{cases} 2 a_{22} + b_{22} = 2 \Rightarrow 2 a_{22} + b_{22} = 2 \\ 2 b_{22} + a_{22} = 1 \Rightarrow 2 a_{22} + 4 b_{22} = 2 \end{cases}$ $\Rightarrow 3 b_{22} = 0 \Rightarrow b_{22} = 0 \Rightarrow a_{22} = 1$ ${\begin{cases} 2 a_{12} + b_{21} = 5 \Rightarrow 2 a_{12} + b_{21} = 5 \\ 2 b_{21} + a_{12} = 4 \Rightarrow 2 a_{12} + 4 b_{21} = 8 \end{cases}$ $3 b_{21} = 3 \Rightarrow b_{21} = 1 \Rightarrow a_{12} = 2$ ${\begin{cases} 2 a_{21} + b_{12} = 10 \Rightarrow 2 a_{21} + b_{12} = 10 \\ 2 b_{12} + a_{21} = 8 \Rightarrow 2 a_{21} + 4 b_{12} = 16 \end{cases}$ $3 b_{12} = 6 \Rightarrow b_{12} = 2 \Rightarrow a_{21} = 4$ $A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = [\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}]$ $B = [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] = [\begin{matrix} 0 & 2 \\ 1 & 0 \end{matrix}]$
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