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Question: Answered & Verified by Expert
In matrix notation, if the system of equations $\left[\begin{array}{c}1 \\ -1 \\ 2\end{array}\right]\left[\begin{array}{ll}1-1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}5 \\ -5 \\ 10\end{array}\right]$ has infinite number of solution, then all these solutions lie on
MathematicsThree Dimensional GeometryTS EAMCETTS EAMCET 2021 (06 Aug Shift 2)
Options:
  • A a line on XY-plane
  • B a plane not parallel to any of the coordinate
    planes.
  • C the YZ-plane.
  • D the ZX-plane.
Solution:
1468 Upvotes Verified Answer
The correct answer is: a plane not parallel to any of the coordinate
planes.
$\begin{aligned}
& \text { Given, }\left[\begin{array}{c}
1 \\
-1 \\
2
\end{array}\right]\left[\begin{array}{ll}
1-1 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
5 \\
-5 \\
10
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & -1 & 2 \\
-1 & 1 & -2 \\
2 & -2 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
5 \\
-5 \\
10
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
x-y+2 z \\
-x+y-2 z \\
2 x-2 y+4 z
\end{array}\right]=\left[\begin{array}{c}
5 \\
-5 \\
10
\end{array}\right]
\end{aligned}$
i.e.
$\begin{gathered}
x-y+2 z=5 \\
-x+y-2 z=-5 \\
2 x-2 y+4 z=10
\end{gathered}$
From Eqs. (i), (ii) and (iii), we observe that the solution lies on a plane not parallel to any of the coordinate plane.

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