Let S be the set of all column matrices [ l b_1 b_2 b_3 ] such th…

MCQ

Let $S$ be the set of all column matrices $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ such that $b_1, b_2, b_3 \in R$ and the system of equations (in real variables)$-x+2 y+5 z=b_1$
$2 x-4 y+3 z=b_2$
$x-2 y+2 z=b_3$
has at least one solution. Then, which of the following system$(s) ($in real variables$)$ has $($have$)$ at least one solution for each$\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$ ?
$(A) \ x+2 y+3 z=b_1, 4 y+5 z=b_2$ and $x+2 y+6 z=b_3$
$(B) \ x+y+3 z=b_1, 5 x+2 y+6 z=b_2$ and $-2 x-y-3 z=b_3$
$(C) \ -x+2 y-5 z=b_1, 2 x-4 y+10 z=b_2$ and $x-2 y+5 z=b_3$
$(D) \ x+2 y+5 z=b_1, 2 x+3 z=b_2$ and $x+4 y-5 z=b_3$

✓
$\text{A,C,D}$
B
$\text{A,C,B}$
C
$\text{A,C}$
D
$\text{A,D}$

✓

Answer

Correct option: A.

$\text{A,C,D}$

For atleast one solution, either $\Delta \neq 0$ or $\Delta=\Delta_1=\Delta_2=\Delta_3=0$.
$\begin{aligned} \Delta & =\left|\begin{array}{ccc}-1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2\end{array}\right|=0\end{aligned} $
$\Delta_1=\left|\begin{array}{ccc}b_1 & 2 & 5 \\ b_2 & -4 & 3 \\ b_3 & -2 & 2\end{array}\right|=0$
$\Rightarrow b_1+7 b_2-13 b_3=0 $
$\Delta_2=\left|\begin{array}{ccc}-1 & b_1 & 5 \\ 2 & b_2 & 3 \\ 1 & b_3 & 2\end{array}\right|=0$
$\Rightarrow b_1+7 b_2-13 b_3=0$
Also, $\Delta_3=0$
For option $(A)$, $\Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6\end{array}\right|=12 \neq 0$, so unique solution.
For option $(B)$, $\Delta=\left|\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right|=0, \Delta_1=0, \Delta_2=\left|\begin{array}{ccc}1 & b_1 & 3 \\ 5 & b_2 & 6 \\ -2 & b_3 & -3\end{array}\right|=3\left(b_1+b_2+3 b_3\right) \neq 0$
So no solution.
For option $( C ), \Delta=\left|\begin{array}{ccc}-1 & 2 & -5 \\ 2 & -4 & 10 \\ 1 & -2 & 5\end{array}\right|=0$
Also, $\Delta_1=\Delta_2=\Delta_3=0$. So, infinitely many solution.
For option $(D)$, $\Delta=\left|\begin{array}{ccc}1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & -5\end{array}\right|=54 \neq 0$, so unique solution.
Hence $(A), (C), (D)$ are correct.