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(C) Let $A = \begin{bmatrix} x & x & x \ x & x^2 & x \ x & x & x+1 \end{bmatrix}$.For the rank of $A$ to be $1$, all minors of order $2$ must be zer

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$x=0$

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(C) Let $A = \begin{bmatrix} x & x & x \ x & x^2 & x \ x & x & x+1 \end{bmatrix}$.For the rank of $A$ to be $1$, all minors of order $2$ must be zero.Consider the minor formed by the first two rows and first two columns: $\begin{vmatrix} x & x \ x & x^2 \end{vmatrix} = x^3 - x^2 = x^2(x-1)$.For this to be $0$, $x=0$ or $x=1$.Case $1$: If $x=1$, $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 2 \end{bmatrix}$. The minor $\begin{vmatrix} 1 & 1 \ 1 & 2 \end{vmatrix} = 2-1 = 1 
eq 0$. Thus, the rank is at least $2$. So $x=1$ is not the solution.Case $2$: If $x=0$, $A = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$. All minors of order $2$ are zero, and there is at least one non-zero element (the $1$ at $A_{33}$). Thus, the rank is $1$.Therefore, the only possible value is $x=0$.

If the rank of the matrix $\begin{bmatrix} x & x & x \\ x & x^2 & x \\ x & x & x+1 \end{bmatrix}$ is $1$, then:

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