Matrices and Determinants 4 Question 16
17. The number of value of $k$, for which the system of equation
$$ (k+1) x+8 y=4 y \quad \Rightarrow \quad k x+(k+3) y=3 k-1 $$
(2013 Main)
has no solution, is
(a) infinite
(b) 1
(c) 2
(d) 3
Show Answer
Answer:
Correct Answer: 17. (d)
Solution:
- Given equations can be written in matrix form $A X=B$
where, $A=\begin{array}{cc}k+1 & 8 \ k & k+3\end{array}, X=\frac{x}{y}$ and $B=\frac{4 k}{3 k-1}$
For no solution, $|A|=0$ and $(\operatorname{adj} A) B \neq 0$
Now, $|A|=\begin{array}{cc}k+1 & 8 \ k & k+3\end{array}=0$
$$ \begin{array}{rlrl} & \Rightarrow & \left(k^{2}+1\right)(k+3)-8 k & =0 \\ & k^{2}+4 k+3-8 k & =0 \\ & k^{2}-4 k \times 3 & =0 \\ \Rightarrow & (k-1)(k-3) & =0 \\ & & k=1, k & =3, \end{array} $$
$$ \text { Now } \quad \operatorname{adj} A=\begin{array}{ll} k+3 & -8 \\ -k & k+1 \end{array} $$
Now, $\quad(\operatorname{adj} A) B=\begin{array}{ccc}k+3 & -8 & 4 k \ -k & k+1 & 3 k+1\end{array}$
$$ \begin{aligned} = & (k+3)(4 k)-8(3 k-1) \\ & -4 k^{2}+(k+1)(3 k-1) \\ = & 4 k^{2}-12 k+8 \\ & -k^{2}+2 k-1 \end{aligned} $$
Put $k=1$
$$ (\operatorname{adj} A) B=\begin{array}{r} 4-12+8 \\ -1+2-1 \end{array}=\begin{aligned} & 0 \\ & 0 \end{aligned} \text { not true } $$
Put $k=3$
$$ (\operatorname{adj} A) B=\begin{gathered} 36-36+8 \\ -9+6-1 \end{gathered}={ } _{-4}^{8} \neq 0 \text { true } $$
Hence, required value of $k$ is 3 .
Alternate Solution
Condition for the system of equations has no solution is
$$ \begin{aligned} \frac{a _1}{a _2} & =\frac{b _1}{b _2} \neq \frac{c _1}{c _2} \\ \therefore \quad \frac{k+1}{k} & =\frac{8}{k+3} \neq \frac{4 k}{3 k-1} \end{aligned} $$
Take $\frac{k+1}{k}=\frac{8}{k+3}$
$\Rightarrow \quad k^{2}+4 k+3=8 k$
$\Rightarrow \quad k^{2}-4 k+3$
$\Rightarrow \quad(k-1)(k-3)=0$
$$ k=1,3 $$
If $k-1$, then $\frac{8}{1+3} \neq \frac{4.1}{2}$, false
And, if $k=3$, then $\frac{8}{6} \neq \frac{4.3}{9-1}$, true
Therefore, $k=3$
Hence, only one value of $k$ exist.
$$ \begin{array}{ll} x & 1 \end{array} $$
