Question

If the system of equations $(k+1)x+8y=4k$ and $kx+(k+3)y=3k-1$ has infinitely many solutions, then $k$ is equal to

• A. $0$
• B. $1$
• C. $3$
• D. $-3$

Answer 1

Answer: (B)

Given system of equation is $(k+1)x+8y=4k$ and $kx+(k+3)y=3k-1.$
$\therefore$ For in fined many solution, $\left|\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right|=0$
$(k+1)(k+3)-8k=0$
$\Rightarrow$ $k^2+4k+3-8k=0$
$\Rightarrow$ $k^2-4k+3=0$
$\Rightarrow$ $k=\frac{4\pm \sqrt{16-12}}{2(1)}$ $=\frac{4\pm 2}{2}=3,1$ Now, $adj(A)=\left[\begin{array}{cc}k+3& -8\\ -k& k+1\end{array}\right]$
Now, $(adjA)B=\left[\begin{array}{cc}k+3& -8\\ -k& k+1\end{array}\right]\left[\begin{array}{c}4k\\ 3k-1\end{array}\right]$
$=\left[\begin{array}{c}4k^2+12k-24k+8\\ -4k^2+3k^2+2k-1\end{array}\right]$
$=\left[\begin{array}{c}4k^2-12k+8\\ -k^2+2k-1\end{array}\right]$
when $k=1,$ $(adjA)B=\left[\begin{array}{c}4-12+8\\ -1+2-1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$
satisfies when $k=3,$
$(adjA)B=\left[\begin{array}{c}36-36+8\\ -9+6-1\end{array}\right]=\left[\begin{array}{c}8\\ -4\end{array}\right]\ne 0$ not satisfies
Hence, $k=1$ is the required solution.