The number of values of k for which the system of equations

$(k+1)x+8y = 4k $

$kx+(k+3)y = 3k-1$

has infinitely many solutions, is

1

The correct answer is option (2) : 1

The given system of equations will have infinitely many solutions, if

$\begin{bmatrix}k+1 & 8 \\k & k+3 \end{bmatrix}=0, \begin{bmatrix}k+1 & 4k \\k & 2k-1 \end{bmatrix}=0 $ and $\begin{bmatrix}4k & 8 \\3k-1 & k+3 \end{bmatrix}=0$

$⇒k^2-4k + 3=0, -k^2 + 2k-1=0$ and $4k^2 - 12k + 8 = 0 $

$⇒(k-1) (k-3) =0, (k-1)^2=0, (k-1) (k-2) = 0 $

$⇒k= 1$