Answer:
Option B
Explanation:
Given equations can be written in matrix form
AX=B
Where, A= $\begin{bmatrix}k+1 & 8 \\k& k+3 \end{bmatrix}=\begin{bmatrix}x \\y \end{bmatrix} and \begin{bmatrix}4k \\3k-1 \end{bmatrix}$
For no solution , |A|= 0 and (adj A)B≠ 0
nOW, |A|= $\begin{bmatrix}k+1 & 8 \\k& k+3 \end{bmatrix}=0$
$\Rightarrow$ (k+1)(k+3)-8K=0
$\Rightarrow$ k2+4k+3-8k=0
$\Rightarrow$ k2-4k+3=0
$\Rightarrow$ (k-1) (k-3)=0
$\Rightarrow$ k=1,k=3
Now, adj A= $\begin{bmatrix}k+3 & -8 \\-k & k+1 \end{bmatrix}$
Now, (adj A)B= $\begin{bmatrix}k+3 & -8 \\-k & k+1 \end{bmatrix}$ $\begin{bmatrix}4k \\3k-1 \end{bmatrix}$
= $\begin{bmatrix}(k+3)(4k)-8(3k-1) \\-4k^{2}+(k+1)(3k-1) \end{bmatrix}$
= $\begin{bmatrix}4k^{2}-12k+8 \\-k^{2}+2k-1 \end{bmatrix}$
Put k=1
(adj A)B= $\begin{bmatrix}4-12-8\\-1+2-1 \end{bmatrix}=\begin{bmatrix}0 \\0 \end{bmatrix}$ not true
Put k=3
(adj A) B= $\begin{bmatrix}36-36+8 \\-9+6-1 \end{bmatrix}=\begin{bmatrix}8 \\-4 \end{bmatrix}\neq0$ true
Hence , required value of k is 3
only onne value of k exist
