Given matrix is
$A=\begin{pmatrix}
1 & 0 & 0\\
0 & a & 0\\
0 & 0 & b
\end{pmatrix}$
Since it is a diagonal matrix, its eigenvalues are the diagonal entries, that is, $1$, $a$, $b$.
Now the question says that an arbitrary vector $X$ is an eigenvector of $A$.
Let
$X=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$
Then for $X$ to be an eigenvector, we must have
$AX=cX$
for some scalar $c$.
Now take $c=1$. Then
$AX=X$
So,
$\begin{pmatrix}
1 & 0 & 0\\
0 & a & 0\\
0 & 0 & b
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\x_3
\end{pmatrix}
=
\begin{pmatrix}
x_1\\x_2\\x_3
\end{pmatrix}$
This gives
$\begin{pmatrix}
x_1\\
ax_2\\
bx_3
\end{pmatrix}
=
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}$
Comparing both sides, we get
$x_1=x_1$
$ax_2=x_2$
$bx_3=x_3$
Since $X$ is arbitrary, $x_2$ and $x_3$ can be any values.
So we must have
$a=1$
and
$b=1$
Hence,
$(a,b)=(1,1)$
So the correct answer is $\boxed{(a,b)=(1,1)}$
