Show that two C*-algebras are morita equivalent

Let $\displaystyle A$ be a C*-algebra, I would like to show that $\displaystyle A$ is Morita equivalent to $\displaystyle M_l(A)$

Here follows a description of my plan to show this:

Define $\displaystyle F(A):=\{(a_1,\dots,a_l):a_i\in A\}$

let $\displaystyle \overline{a}:=(a_1,\dots,a_l)$ be the row matrix, then we can define the following inner products

$\displaystyle _{A}\langle \overline{a},\overline{b}\rangle:=\overline{a}\ove rline{b}^*$

$\displaystyle \langle \overline{a},\overline{b}\rangle_{M_l(A)}:=\overli ne{a}^*\overline{b}$

The above choice of inner product yields the desired result which we need to have an imprimitivity bimodule, namely

$\displaystyle _{A}\langle \overline{a},\overline{b}\rangle \overline{c}=\overline{a}\overline{b}^*\overline{c }=\overline{a}\langle \overline{b},\overline{c}\rangle_{M_l(A)}$

I can show that $\displaystyle F(A)$ in complete in the norm induced by the $\displaystyle A$-inner product. But not for the norm induced by the $\displaystyle M_l(A)$-inner product.

I can also show that $\displaystyle I_{A}:=\text{span}\{_{A}\langle \overline{a},\overline{b}\rangle:\overline{a},\ove rline{b}\in A\}$ is dense in $\displaystyle A$. Is it possible to show that

$\displaystyle I_{M_{l}(A)}:=\text{span}\{\langle \overline{a},\overline{b}\rangle_{M_l(A)}:\overlin e{a},\overline{b}\in A\}$ is dense in $\displaystyle M_l(A)$?

Or is this idea totally wrong?