A) \[b+c\]
B) \[a+d\]
C) \[ab+cd\]
D) zero
Correct Answer: D
Solution :
Given ,\[A=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} {{a}^{2}}+bc & ab+bd \\ ac+dc & bc+{{d}^{2}} \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}-(a+d)A+kI=0\] \[\Rightarrow \]\[\left[ \begin{matrix} {{a}^{2}}+bc & ab+bd \\ ac+dc & dc+{{d}^{2}} \\ \end{matrix} \right]-\left[ \begin{matrix} {{a}^{2}}+ad & ab+bd \\ ac+dc & ad+{{d}^{2}} \\ \end{matrix} \right]\] \[+\left[ \begin{matrix} k & 0 \\ 0 & k \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\left[ \begin{matrix} bc-ad+k & 0 \\ 0 & bc-ad+k \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\] On equating, we get \[bc=ad+k=0\] \[\Rightarrow \] \[k=ad-bc\] Also, \[\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[ad-bc=0\] \[\therefore \] From Eq. (i), \[k=0\]
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