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The System of Linear Equations: X + Y + Z = 2 2x + Y − Z = 3 3x + 2y + Kz = 4 Has a Unique Solution If (A) K ≠ 0 (B) −1 < K < 1 (C) −2 < K < 2 (D) K = 0 - Mathematics

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Question

The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if

Options

  • k ≠ 0

  • −1 < k < 1

  • −2 < k < 2

  •  k = 0

MCQ
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Solution

(a) k ≠ 0

\[\text{ For a unique solution, }\left| A \right|\neq 0.\]
The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}1 & 1 & 1 \\ 2 & 1 & - 1 \\ 3 & 2 & k\end{bmatrix}\neq0\]
\[ \Rightarrow 1\left( k + 2 \right) - 1\left( 2k + 3 \right) + 1\left( 4 - 3 \right)\neq0\]
\[ \Rightarrow k + 2 - 2k - 3 + 1\neq 0\]
\[ \Rightarrow k \neq 0\]
So, the given system of equations has a unique solution ifkis not equal to 0.

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Chapter 8: Solution of Simultaneous Linear Equations - Exercise 8.4 [Page 22]

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RD Sharma Mathematics [English] Class 12
Chapter 8 Solution of Simultaneous Linear Equations
Exercise 8.4 | Q 5 | Page 22

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