Determinants Matrices Question 15
Question: For what values of k, does the system of linear equation $ x+y+z=2, $
$ 2x+y-z=3, $
$ 3x+2y+kz=4 $ have a unique solution-
Options:
A) $ k=0 $
B) $ -1<k<1 $
C) $ -2<k<2 $
D) $ k\ne 0 $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] The given system of equations is $ x+y+z=2 $
… (i) $ 2x+y-z=3 $
… (ii) and $ 3x+2y+kz=4 $
… (iii) This system has a unique solution if $ \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \\ \end{vmatrix}\ne 0 $
Applying $ C_2\to C_2-C_1 $ and $ C_3\to C_3-C_1 $ We get $ \begin{vmatrix} 1 & 0 & 0 \\ 3 & -1 & -3 \\ 3 & -1 & k-3 \\ \end{vmatrix}\ne 0 $
$ \Rightarrow -1(k-3)-3\ne 0 $ or $ -k+3-3\ne 0\Rightarrow k\ne 0 $
BETA
