Question

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

• A. infinite
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

Given equations can be written in matrix from AX = B
where, $A=\left[\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]andB=\left[\begin{array}{c}4k\\ 3k-1\end{array}\right]$
For no solution, |A| = 0 and (adj A) B $\ne$ 0
Now, $|A|=\left[\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right]=0$
$\Rightarrow (k+1)(k+3)-8k=0k^2+4k+3-8k=0\Rightarrow k^2-4k+3=0\Rightarrow (k-1)(k-3)=0\Rightarrow k=1,k=3,$
Now $adjA=\left[\begin{array}{cc}k+3& -8\\ -k& k+1\end{array}\right]$
Now, $\left(adjA\right)B=\left[\begin{array}{cc}k+3& -8\\ -k& k+1\end{array}\right]\left[\begin{array}{c}4k\\ 3k+1\end{array}\right]$
$=\left[\begin{array}{c}(k+3)\left(4k\right)-8(3k-1)\\ -4k^2+(k+1)(3k-1)\end{array}\right]=\left[\begin{array}{c}4k^2-12k+8\\ -k^2+2k-1\end{array}\right]$
Put k = 1
$\left(adjA\right)B=\left[\begin{array}{c}4-12+8\\ -1+2-1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$ not true
Put k = 3
$\left(adjA\right)B=\left[\begin{array}{c}36-36+8\\ -9+6-1\end{array}\right]=\left[\begin{array}{c}8\\ -4\end{array}\right]\ne 0true$
Hence, required value of k is 3.
Alternate Solution
Condition for the system of equations has no solution is
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\therefore \frac{k+1}{k}=\frac{8}{k+3}\ne \frac{4k}{3k-1}Take\frac{k+1}{k}=\frac{8}{k+3}\Rightarrow k^2+4k+3=8k\Rightarrow k^2-4k+3=0\Rightarrow (k-1)(k-3)=0$
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.