### Determinants Question 5

#### Question 5 - 30 January - Shift 2

For $\alpha, \beta \in \mathbb{R}$, suppose the system of linear equations

$\mathbf{x}-y+z=5$

$2 x+2 y+\alpha z=8$

$3 x-y+4 z=\beta$

has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of

(1) $x^{2}-10 x+16=0$

(2) $x^{2}+18 x+56=0$

(3) $x^{2}-18 x+56=0$

(4) $x^{2}+14 x+24=0$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: System of equations with 3 variables, consistency of solutions, Evaluation of the Determinant

$ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4\end{vmatrix} =0 ; 8+\alpha-2(-4+1)+3(-\alpha-2)=0$

$8+\alpha+6-3 \alpha-6=0$

$\alpha=4$

Also, for infinite solution,

$ \begin{aligned} & \begin{vmatrix} 1 & -1 & 5 \\ 2 & 2 & 8 \\ 3 & -1 & \beta\end{vmatrix}=0 \\ & \Rightarrow 5(-2-6) \cdot 8(-1+3)+\beta(2+2)=0 \\ & \Rightarrow 4 \beta=40+16=56 \\ & \Rightarrow \beta=14 \\ & \therefore \alpha+\beta=4+14=18 \\ & \& \alpha \beta=4 \times 14=56 \end{aligned} $

$\therefore$ Quadratic equation whose roots are $\alpha \& \beta$ is $ x^2-18 x+56=0 $