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:

  1. 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).



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