A) \[a=4,b=1\]
B) \[a=1,b=4\]
C) \[a=0,b=4\]
D) \[a=2,b=4\]
Correct Answer: B
Solution :
Given, \[A=\left[ \begin{matrix} 1 & -1 \\ 2 & -1 \\ \end{matrix} \right],\,\,B=\left[ \begin{matrix} a & 1 \\ b & -1 \\ \end{matrix} \right]\] Þ \[A+B=\left[ \begin{matrix} 1+a & 0 \\ 2+b & -2 \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} 1 & -1 \\ 2 & -1 \\ \end{matrix} \right]\,\,\left[ \begin{matrix} 1 & -1 \\ 2 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right]\] \[{{B}^{2}}=\left[ \begin{matrix} a & 1 \\ b & -1 \\ \end{matrix} \right]\,\,\,\left[ \begin{matrix} a & 1 \\ b & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+b & a-1 \\ ab-b & b+1 \\ \end{matrix} \right]\] \[{{A}^{2}}+{{B}^{2}}=\left[ \begin{matrix} {{a}^{2}}+b-1 & a-1 \\ ab-b & b \\ \end{matrix} \right]\] Also, \[{{(A+B)}^{2}}=\left[ \begin{matrix} 1+a & 0 \\ 2+b & -2 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1+a & 0 \\ 2+b & -2 \\ \end{matrix} \right]\] \[{{(A+B)}^{2}}=\left[ \begin{matrix} {{(1+a)}^{2}} & \,\,\,\,0 \\ (2+b)(1+a)-2(2+b) & \,\,\,\,4 \\ \end{matrix} \right]\] \ \[{{(A+B)}^{2}}={{A}^{2}}+{{B}^{2}}\] Þ \[\left[ \begin{matrix} {{(1+a)}^{2}} & 0 \\ \,(2+b)(a-1) & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+b-1 & a-1 \\ ab-b & b \\ \end{matrix} \right]\] By equating, \[a-1=0\Rightarrow a=1\] and \[b=4\].
You need to login to perform this action.
You will be redirected in
3 sec
