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 $AX=B$
where, $A=\left[\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right]=0,X=\left[\frac{x}{y}\right]$ and $B=\frac{4k}{3k-1}$
For no solution, $|A|=0$ and $(adjA)B\ne 0$
Now $|A|=\left[\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right]=0\Rightarrow (k^2+1)(k+3)-8k=0$
$k^2+4k+3-8k=0$
$\Rightarrow k^2-4k\times 3=0$
$\Rightarrow (k-1)(k-3)=0\Rightarrow k=1,k=3$
Now $adjA=\left[\begin{array}{cc}k+3& -8\\ -k& k+1\end{array}\right]=0$
Now $(adjA)B=\left[\begin{array}{cc}k+3& -8\\ -k& k+1\end{array}\right]=\left[\begin{array}{c}4k+3\\ 3k+1\end{array}\right]=0$
$=\left[\begin{array}{cc}(k+3)(4k)& -8(3k-1)\\ -4k^2+& (k+1)(3k-1)\end{array}\right]$
$=\left[\begin{array}{c}4k^2-12k+8\\ -k^2+2k-1\end{array}\right]$
Put $k=1$
$(adjA)B=\left[\begin{array}{c}4-12+8\\ -1+2-1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$ not true
Put $k=3$
$(adjA)B=\left[\begin{array}{c}36-36+8\\ -9+6-1\end{array}\right]=\left[\begin{array}{c}8\\ -4\end{array}\right]\ne 0$ 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.