For what value of k the system of the following linear equations has infinitely many solutions?kx+4y=k-4 and 16x+ky=k.

Given,

kx + 4 y = k - 4

16 x + ky = k

.
Here,

a_{1} = k, b_{1} = 4, c_{1} = - (k - 4)

.
And

a_{2} = 16, b_{2} = k, c_{2} = - k

.
The given equations are consistent if,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

⇒

\frac{k}{16} = \frac{4}{k} = \frac{k - 4}{k}

⇒

\frac{k}{16} = \frac{4}{k}

⇒

k^{2} = 64

⇒

k = \pm 8

Also,

\frac{4}{k} = \frac{k - 4}{k}

.
⇒

4 k = k^{2} - 4 k

⇒

k^{2} - 8 k = 0

⇒

k (k - 8) = 0

⇒

k = 0

k = 8

But

k = 0

is not possible.
Therefore,

k = 8

is the correct value for infinitely many solutions.
Hence, the correct option is 2.

For what value of $k$ the system of the following linear equations has infinitely many solutions?

$kx + 4 y = k - 4$ and $16 x + ky = k$ .