Answer
False
Work Step by Step
Consider $E_1=\begin{bmatrix}
1 & a\\0&1
\end{bmatrix}$ and $E_2=\begin{bmatrix}
1 & 0\\b&1
\end{bmatrix}$ are $n \times n$ elementary matrices of the same type, then we obtain:
$E_1E_2=\begin{bmatrix}
1 & a\\0&1
\end{bmatrix}\begin{bmatrix}
1 & 0\\b&1
\end{bmatrix}=\begin{bmatrix}
1+a & a+b\\1&1
\end{bmatrix}\\
E_2E_1=\begin{bmatrix}
1 & 0\\b&1
\end{bmatrix}\begin{bmatrix}
1 & a\\0&1
\end{bmatrix}=\begin{bmatrix}
1 & a+b\\1&1+a
\end{bmatrix}$
We can see that $E_1E_2$ and $E_2E_1$ are not the same.
Hence, the statement is false.
