A) \[0,12,-12\]
B) \[0,1,-1\]
C) \[0,4,-4\]
D) \[0,5,-5\]
Correct Answer: A
Solution :
We have, \[A=\left[ \begin{matrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \\ \end{matrix} \right]\] Since, A is singular \[\therefore \] \[|A|=0\] \[\Rightarrow \] \[0-x[{{x}^{2}}-0]+16[9x-0]=0\] \[\Rightarrow \] \[-{{x}^{3}}+144x=0\] \[\Rightarrow \] \[x({{x}^{2}}-144)=0\] \[\Rightarrow \] \[x=0,\] \[x=\pm 12\]
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