Answer
See below
Work Step by Step
We are given $\begin{bmatrix}
a & b\\
b & c
\end{bmatrix}$
With $\det(A- \lambda I)$
we obtain $\begin{vmatrix}
a-\lambda & b\\
b & c-\lambda
\end{vmatrix}=0 \\
\rightarrow \frac{(a+c) \pm \sqrt (a-c)^2+4b^2}{2}$
Since $A$ has repeated $\lambda$
then $(a-c)^2+4b^2=0\\
\rightarrow a=c\\
b=0$
with $a,b,c \in R$
Hence, $\begin{bmatrix}
a & 0\\
0 & a
\end{bmatrix}$
Hence, $A$ is a scalar matrix.
