Question

If $A$ and $B$ are the two real values of $k$ for which the system of equations $x+2 y+z=1$, $x+3 y+4 z=k . x+5 y+10 z=k^{2}$ is consistent, then $A+B=$

• A. 3
• B. 4
• C. $\underline{5}$
• D. 7

Answer 1

Answer: (A)

Given system of equation is,
$ x+2 y+z=1 $
$ x+3 y+4 z=k $
$ x+5 y+10 z=k^{2} $
$\therefore \, D=\begin{vmatrix}1 & 2 & 1 \\1 & 3 & 4 \\1 & 5 & 10\end{vmatrix} $
$= 1(30-20)-2(10-4)+1(5-3)=10-12+2=0$
Since, $D=0$
$\therefore $ Given system of equation is consistent.
Therefore, $D_{1}=0$
$D_{1}=\begin{vmatrix}1 & 2 & 1 \\k & 3 & 4 \\k^{2} & 5 & 10\end{vmatrix}$
$\Rightarrow \, 1(30-20)-2\left(10 k-4 k^{2}\right)+\left(5 k-3 k^{2}\right)=0$
$\Rightarrow \,10-20 k+8 k^{2}+5 k-3 k^{2}=0$
$\Rightarrow \, 5 k^{2}-15 k+10=0$
$\Rightarrow \, k^{2}-3 k+2=0$
$\Rightarrow \, (k-2)(k-1)=0$
$\Rightarrow \, k=2,1$
Hence, the real values of $k$ i.e.
$A=2$ and $B=1$
$\therefore \, A+B=2+1=3$