Question

The number of value of $k$, for which the system of equation $(k+1)x+8y=4y \, \, \Rightarrow \, \, \, kx+(k+3)y=3k-1$ has no solution, is

• A. infinite
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

Given equations can be written in matrix form $A X = B$
where, $A =\begin {bmatrix}k+1 & 8 \\k & k+3 \end {bmatrix}=0 , X=\bigg[ \frac{x}{y}\bigg] $ and $B=\frac{4k}{3k-1}$
For no solution, $|A| =0$ and $(adj A) B \ne 0 $
Now $ |A| =\begin {bmatrix}k+1 & 8 \\k & k+3 \end {bmatrix}=0 \Rightarrow (k^2+1)(k+3)-8k=0$
$ k^2+4k+3-8k=0$
$\Rightarrow k^2-4k \times 3 =0$
$\Rightarrow (k-1)(k-3)=0 \Rightarrow k=1,k=3$
Now $ adj A =\begin {bmatrix}k+3 & -8 \\-k & k+1 \end {bmatrix}=0 $
Now $ (adj \ A)B=\begin {bmatrix}k+3 & -8 \\-k & k+1 \end {bmatrix}= \begin {bmatrix}4k+3 \\3k+1 \end {bmatrix}=0 $
$ =\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 $k=3$
$(adj A) B = \begin {bmatrix}36-36+8 \\-9+ 6-1 \end {bmatrix}=\begin {bmatrix} 8 \\-4 \end {bmatrix} \ne 0$ true
Hence, required value of $k$ is $3$.
Alternate Solution
Condition for the system of equations has no solution is
$ \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\therefore$ Take $ \frac{k+1}{k}=\frac{8}{k+3} \ne \frac{4k}{3k-1} $
Take $ \frac{k+1}{k}=\frac{8}{k+3} \Rightarrow k^2+4k+3=8k$
$\Rightarrow k^2- 4k+3 \Rightarrow (k -1)(k-3) = 0$
$k = 1,3$
If $k - 1$, then $\frac{8}{1+3}\ne \frac{4.1}{2}, $ false
And, if $k =3$, then $\frac{8}{6} \ne \frac{4.3}{9-1} $ true
Therefore, $k=3 $
Hence, only one value of $k$ exist.