The rank of the matrix 4 2 ( 1 - x ) 5 k 1 6 3 ( 1 + x ) is 1 , then

Given that Rank of the matrix $\left[\begin{array}{ccc}4& 2& (1-x)\\ 5& k& 1\\ 6& 3& (1+x)\end{array}\right]$ is $1$.

We know that, Rank of Matrix is number of non-zero rows of a matrix in its echelon form.

Applying $R_3\to R_1+R_3$

$\Rightarrow \left[\begin{array}{ccc}4& 2& (1-x)\\ 5& k& 1\\ 10& 5& 2\end{array}\right]$

Now, Applying $R_3\to R_3-2R_2$

$\Rightarrow \left[\begin{array}{ccc}4& 2& (1-x)\\ 5& k& 1\\ 0& 5-2k& 0\end{array}\right]$

Now, Applying $R_2\to R_2-\frac{5}{4}{R}_1$

$\Rightarrow \left[\begin{array}{ccc}4& 2& (1-x)\\ 0& k-\frac{5}{2}& \frac{5x-1}{4}\\ 0& 5-2k& 0\end{array}\right]$

Rank of Matrix is $1$

$\Rightarrow k-\frac{5}{2}=0, \frac{5x-1}{4}=0, 5-2k=0$

$\therefore k=\frac{5}{2}, x=\frac{1}{5}$