A) \[-3\]
B) \[3\]
C) \[0\]
D) for any value of\[b\]
Correct Answer: D
Solution :
Key Idea: If any square matrix is singular, then the value of determinant is zero. Let \[A=\left[ \begin{matrix} 5 & 10 & 3 \\ 2 & -4 & 6 \\ -1 & -2 & b \\ \end{matrix} \right]\] Since, \[A\] is singular. \[\therefore \] \[\left| \begin{matrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & b \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[5(-4b+12)-10(-2b+6)+3(4-4)=0\] \[\Rightarrow \]\[-20b+60+20b-60+0=0\] \[\Rightarrow \]\[0=0\] \[\therefore \]The given matrix is singular for any value of\[b\]. Note: For a non-singular matrix, the value of determinant is non-zero.
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