| \[2x+2y+3z=a\] |
| \[3x-y+5z=b\] |
| \[x-3y+2z=c\] |
| where a, b, c are non-zero real numbers, has more than one solution, then |
A) \[b+c-a=0\]
B) \[a+b+c=0\]
C) \[b-c+a=0\]
D) \[b-c-a=0\]
Correct Answer: D
Solution :
We have,\[A=\left[ \begin{matrix} 2 & 2 & 3 \\ 3 & -1 & 5 \\ 1 & -3 & 2 \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]\]and\[B=\left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]\] Here, \[\left| A \right|=0\] Since, the system of linear equations has more than one solution, so (adj A) = 0 \[\Rightarrow \left[ \begin{matrix} 13 & -13 & 13 \\ -1 & 1 & -1 \\ -8 & 8 & -8 \\ \end{matrix} \right]=\left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} \right]\]\[\Rightarrow b=a+c\]
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