Answer
$W$ is a vector subspace of $M_{3,2}$.
Work Step by Step
Let $u=\left[\begin{array}{ll}{a} & {b} \\ {a+b} & {0}\\
{0}&{c}\end{array}\right],v=\left[\begin{array}{ll}{d} & {e} \\ {d+e} & {0}\\
{0}&{f}\end{array}\right],\in W$ and $c'\in R$ where
$W$ the set of all $3\times 2$ of the form
$\left[\begin{array}{ll}{a} & {b} \\ {a+b} & {0}\\
{0}&{c}\end{array}\right]$.
Now,
\begin{align*}
u+v&= \left[\begin{array}{ll}{a} & {b} \\ {a+b} & {0}\\
{0}&{c}\end{array}\right]+\left[\begin{array}{ll}{d} & {e} \\ {d+e} & {0}\\
{0}&{f}\end{array}\right]\\
&=\left[\begin{array}{ll}{a+d} & {b+e} \\ {a+b+d+e} & {0}\\
{0}&{c+f}\end{array}\right]
\end{align*}
which means that $u+v\in W$ and also $c'u= \left[\begin{array}{ll}{c'a} & {c'b} \\ {c'(a+b)} & {0}\\
{0}&{c'c}\end{array}\right]\in W$. Hence, $W$ is a vector subspace of $M_{3,2}$.
