Matrices and Determinants 4 Question 8
8. An ordered pair $(\alpha, \beta)$ for which the system of linear equations
(2019 Main, 12 Jan I)
$ \begin{aligned} & (1+\alpha) x+\beta y+z=2 \\ & \alpha x+(1+\beta) y+z=3 \\ & a x+\beta y+2 z=2 \end{aligned} $
has a unique solution, is
(a) $(2,4)$
(b) $(-4,2)$
(c) $(1,-3)$
(d) $(-3,1)$
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Answer:
Correct Answer: 8. (a)
Solution:
- Given system of linear equations,
$(1+\alpha) x+\beta y+z=2$
$\alpha x+(1+\beta) y+z=3$
$ \alpha x+\beta y+2 z=2 $
has a unique solution, if
$ \left|\begin{array}{ccc} 1+\alpha & \beta & 1 \\ \alpha & (1+\beta) & 1 \\ \alpha & \beta & 2 \end{array}\right| \neq 0 $
Apply $R _1 \rightarrow R _1-R _3$ and $R _2 \rightarrow R _2-R _3$
$ \left|\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \alpha & \beta & 2 \end{array}\right| \neq 0 $
$ \begin{array}{rr} \Rightarrow & 1(2+\beta)-0(0+\alpha)-1(0-\alpha) \neq 0 \\ \Rightarrow & \alpha+\beta+2 \neq 0 \end{array} $
Note that, only $(2,4)$ satisfy the Eq. (i).
