Question

The system of the equations
$x + y + z = 6$
$x + 2y + \alpha z = 5$
x + 2y + 6z = $$ \beta has

• A. Infinitely many solution for $\alpha = 6, \beta= 3$
• B. Infinitely many solution for $\alpha = 6, \beta= 5$
• C. Unique solution for $\alpha = 6, \beta= 5$
• D. No solution for $\alpha = 6, \beta= 5$

Answer 1

Answer: (B)

Infinitely many solution for $\alpha = 6, \beta= 5$

Answer 2

The correct option is (B): Infinitely many solution for $\alpha = 6, \beta= 5$

Let $\triangle$ = $\begin{vmatrix} 1 & 1 & 1\\\ 1 & 2 & a\\\1 & 2& 6 \end{vmatrix}$= $6-\alpha$
$\Longrightarrow$ for \alpha$$ \neq6 system has unique solution
Now, when $\alpha=6$
\triangle_1 =$$\begin{vmatrix} 6 & 1 & 1\\\ 5 & 2 & 6\\\\\beta & 2& 6 \end{vmatrix}$$=0-(30-6\beta)+(10-2\beta)=(4\beta-5)
\triangle_2$$\begin{vmatrix} 1 & 6 & 2\\\ 1 & 5 & 6\\\1 & \beta& 6 \end{vmatrix}=-4(\beta-5)
\triangle_3=$$\begin{vmatrix} 1 & 1 & 1\\\ 1 & 2 & 5\\\1 & 2& \beta \end{vmatrix} =\begin{vmatrix} 1 & 1 & 6\\\ 0 & 1 & -1\\\0 & 1& \beta-6 \end{vmatrix}=\beta-5
Clearly at $\beta$ = 5, $\triangle_i=$ 0 for i = 1, 2, 3
∴ at $\alpha = 6, \beta = 5$ system has infinite solutions.