Question

Assertion :Consider the system of the equations $kx+y+z=1,x+ky+z=k,x+y+kz=k^2$. System of equations has infinite solutions when $k=1$ Reason: If the determinant $\left|\begin{array}{ccc}1& 1& 1\\ k& k& 1\\ k^2& 1& k\end{array}\right|=0$, then $k=-1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Given system of equations $AX=B$
where $A=\left[\begin{array}{ccc}k& 1& 1\\ 1& k& 1\\ 1& 1& k\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}1\\ k\\ k^2\end{array}\right]$
For infinitely many solution,
$D=0$
$\Rightarrow D=\left|\begin{array}{ccc}k& 1& 1\\ 1& k& 1\\ 1& 1& k\end{array}\right|=0$
$\Rightarrow (k-1)(k^2+k-2)=0$
$\Rightarrow (k-1{)}^2(k+2)=0$
$\Rightarrow k=1,-2$
Now, for $k=1$
$D_1=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|$
$\Rightarrow D_1=0$
Also, $D_2=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|=0$
$D_3=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|=0$
So, $D=D_1=D_2=D_3=0$
Hence, at $k=1$ , system has infinitely many solution.
Hence, assertion is correct.
Now, given $\left|\begin{array}{ccc}1& 1& 1\\ k& k& 1\\ k^2& 1& k\end{array}\right|=0$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}0& 0& 1\\ 0& k-1& 1\\ k^2-1& 1-k& k\end{array}\right|=0$
$\Rightarrow {(k-1)}^2(k+1)=0$
$\Rightarrow k=-1,k=1$