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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 $x+4 y-5 z=b_3$
$A,C,D$
$A,C,B$
$A,C$
$A,D$
Solution
For atleast one solution, either $\Delta \neq 0$ or $\Delta=\Delta_1=\Delta_2=\Delta_3=0$.
$\begin{aligned} \Delta & =\left|\begin{array}{ccc}-1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2\end{array}\right|=0 \\ \Delta_1 & =\left|\begin{array}{ccc}b_1 & 2 & 5 \\ b_2 & -4 & 3 \\ b_3 & -2 & 2\end{array}\right|=0 \Rightarrow b_1+7 b_2-13 b_3=0 \\ \Delta_2 & =\left|\begin{array}{ccc}-1 & b_1 & 5 \\ 2 & b_2 & 3 \\ 1 & b_3 & 2\end{array}\right|=0 \quad \Rightarrow b_1+7 b_2-13 b_3=0\end{aligned}$
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.
Similar Questions
Let $p, q, r$ be nonzero real numbers that are, respectively, the $10^{\text {th }}, 100^{\text {th }}$ and $1000^{\text {th }}$ terms of a harmonic progression. Consider the system of linear equations
$x+y+z=1$
$10 x+100 y+1000 z=0$
$q r x+p r y+p q z=0$.
| $List-I$ | $List-II$ |
| ($I$) If $\frac{q}{r}=10$, then the system of linear equations has | ($P$) $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution |
| ($II$) If $\frac{ p }{ r } \neq 100$, then the system of linear equations has | ($Q$) $x =\frac{10}{9}, y =-\frac{1}{9}, z =0$ as a solution |
| ($III$) If $\frac{p}{q} \neq 10$, then the system of linear equations has | ($R$) infinitely many solutions |
| ($IV$) If $\frac{p}{q}=10$, then the system of linear equations has | ($S$) no solution |
| ($T$) at least one solution |
The correct option is:
