Let $S$ be the set of all column matrices
$
\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}
$
such that $b_1,b_2,b_3 \in \mathbb{R}$ and the system of equations (in real variables)
$$\begin{align*}
-x+2y+5z &=b_1 \nonumber\\
2x-4y+3z &=b_2 \nonumber\\
x-2y+2z &=b_3
\end{align*}$$
has at least one solution.Then, which of the following system(s)(in real variables) has (have) at least one solution for each
$$
\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix} \in S?
$$
A. $x+2y+3z=b_1$, $4y+5z=b_2$ and $x+2y+6z=b_3$
B. $x+y+3z=b_1$, $5x+2y+6z=b_2$ and $-2x-y-3z=b_3$
C. $-x+2y-5z=b_1$, $2x-4y+10z=b_2$ and $x-2y+5z=b_3$
D. $x+2y+5z=b_1$, $2x+3z=b_2$ and $x+4y-5z=b_3$
Can anyone please help me with this problem? I am really clueless how to proceed.
Best Answer
For the Eq. $MX=V$ where $M$ is $3 \times 3$, $X_{3\times 1}$ is unknown vector and $V_{3 \times 1}$ is a fixed (constant). (1) If $V=O$ then $|M|=0$ means many solutions.
(2) If $V\ne O$
(a) then $|M|\ne 0$ means unique solution (planes meeting in a point)
(b) $|M|=0$ means many solutions (identical planes/ two identical planes one intersecting them/ planes meeting in a line)
OR no solution (parallel planes/ two identical and one parallel to them/ two parallel; and third one cutting them/ planes forming a 3D triangle or an open prism)
Here in the description $|M|=0$ and three planes are meeting in a line.
In the options $|A|\ne 0$ and $|D|\ne 0$ so there will be unique solution.
In other two options $|B|=0$. here there are no parallel or identical planes so a prism will be formed so no solution. Next |C|=0$ check there are two parallel planes (no solution).
Finally in (A) and (D) unique solution exists and it is correct option.