Q46 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

Consider the simultaneous linear equations $\beta x+\alpha y-z=-1$, $3 x-\beta y+\alpha z=0 \alpha x+\beta y+z=1$. In the usual notation used in Crammer's rule, given that $\frac{\Delta_1}{\Delta}=-1, \frac{\Delta_2}{\Delta}=1 \cdot \frac{\Delta_3}{\Delta}=2$, then $(\alpha, \beta)=$

  • A
    $(1,2)$
  • Correct answer
    B
    $(2,1)$
  • C
    $(-1,2)$
  • D
    $(1,-2)$

Solution

Here we are given that $$ \mathrm{x}=\frac{\Delta_1}{\Delta}=-1, \quad \mathrm{y}=\frac{\Delta_2}{\Delta}=1, \quad \mathrm{z}=\frac{\Delta_3}{\Delta}=2 $$ From system of linear eqns. $\begin{array}{ll} -\beta+\alpha \times 1-2=-1, & 3 \times(-1)-\beta \times(1)+2 \alpha=0, \alpha \times(-1) \\ +\beta \times(1)+2=1 & \\ \Rightarrow \alpha-\beta=1, & -3-\beta+2 \alpha=0,-\alpha+\beta+2=1 \\ \Rightarrow 2 \alpha-\beta=3, & -\alpha+\beta=-1 \\ \Rightarrow \alpha-\beta=1 & \Rightarrow \alpha-\beta=1 \\ 2 \alpha-\beta=3 & \Rightarrow-\alpha-2 \\ \quad-+ & \Rightarrow \alpha=2 \\ \Rightarrow(\alpha, \beta) & \Rightarrow \beta-\alpha-1=2-1=1 \\ & \Rightarrow \beta=1 \\ & =(2,1) \end{array}$

Hence, the correct answer is option: 'B'.

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