A) 0
B) -4
C) 4 and not 1
D) 1 or 4
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}-4A+10I=A\] \[\Rightarrow \] \[\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]-4\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]\] \[+10\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} -5 & -3-3k \\ 2+2k & -6+{{k}^{2}} \\ \end{matrix} \right]-\left[ \begin{matrix} 4 & -12 \\ 8 & 4k \\ \end{matrix} \right]+\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 1 & 9-3k \\ -6+2k & 4+{{k}^{2}}-4k \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -3 \\ 2 & k \\ \end{matrix} \right]\] \[\Rightarrow \] \[9-3k=-3,\] \[-6+2k=2\] ?.(i) and \[4+{{k}^{2}}-4k=k\] \[\Rightarrow \] \[{{k}^{2}}-5k+4=0\] \[\Rightarrow \] \[(k-4)\,(k-1)=0\] \[\Rightarrow \] \[k=4,1\] But k = 1 is not satisfied the Eq. (i).
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