Unital representations of matrices over a $C^*$-algebra are inflations of unital representations of the $C^*$-algebra this is not a really a question. I recently came across this exercise which I think is very useful and easy to apply in many situations. I am posting this here so that if someone uses this technical result in an answer, there is a reference on MSE for a proof. Also, I am posting this for solution verification; if you see any mistake, please let me know so that I can edit. Here it is:

Let $A$ be a unital C*-algebra, $n\geq1$ and let $(\mathcal{H},\pi)$ be a unital representation of $M_n(A)$. Then there exists a unital representation $(\mathcal{K},\varrho)$ of $A$ so that $(\mathcal{H},\pi)$ is unitarily equivalent to $\varrho^{(n)}:M_n(A)\to\mathbb{B}(\mathcal{K}\oplus\dots\oplus\mathcal{K})$,
$$\varrho^{(n)}[a_{i,j}]:=[\varrho(a_{i,j})].$$

Proof: Let $\{e_{ij}\}$ be the standard matrix units for $M_n(A)$ and set $p_j=\pi(e_{jj})$. These are orthogonal projections in $\mathbb{B}(\mathcal{H})$ and we set $\mathcal{L}_j=p_j(\mathcal{H})$. Observe that $\mathcal{H}=\bigoplus_{j=1}^n\mathcal{L}_j$; indeed, if $\xi\in\mathcal{H}$, then
$$\xi=\pi(1_{M_n(A)})\xi=\sum_{j=1}^n\pi(e_{jj})\xi=\sum_{j=1}^np_j\xi\in\bigoplus_{j=1}^n\mathcal{L}_j$$
The idea is to observe that $\mathcal{L}_j$ are all isomorphic as Hilbert spaces. Let
$$u=\begin{pmatrix}0&1_A\\1_A&0\end{pmatrix}\in M_2(A)$$
which is a self-adjoint unitary matrix. By setting $u_1=u\oplus 1_{M_{n-2}(A)}$, $u_2=1_A\oplus u\oplus 1_{M_{n-3}(A)}$, $u_3=1_{M_2(A)}\oplus u\oplus 1_{M_{n-4}(A)}$ and so on until $u_{n-1}=1_{M_{n-2}(A)}\oplus u$ (which are all written in block form and are all self-adjoint unitaries in $M_n(A)$), we see that
$$e_{22}=u_1\cdot e_{11}\cdot u_1,\;\;e_{33}=u_2\cdot e_{22}\cdot u_2,\dots, e_{nn}=u_{n-1}\cdot e_{n-1,n-1}\cdot u_{n-1}.$$
Since $\pi$ is a unital *-homomomorphism, $v_j:=\pi(u_j)$ are unitary, self-adjoint operators and we have the relations $p_{j+1}=v_jp_jv_j$ for all $j=1,\dots,n-1$. Define $w_j:\mathcal{L}_j\to\mathcal{L}_{j+1}$ by restrcting $v_j$,i.e.
$$w_j(\xi)=v_j(\xi)$$
and note that, since $\xi\in\mathcal{L}_j$ we have that $\xi=p_j(\xi)$ so $v_j\xi=v_jp_j\xi=p_{j+1}v_j\xi\in\mathcal{L}_{j+1}$, so $w_j$ is well-defined. Also, $w_j$ is onto $\mathcal{L}_{j+1}$, since if $\xi\in\mathcal{L}_{j+1}$, then
$$\xi=p_{j+1}\xi=v_jp_jv_j\xi=w_j(p_jv_j\xi)=w_j(\xi')$$
where $\xi'=p_jv_j\xi\in\mathcal{L}_j$. Of course, since $v_j$ is a unitary, $w_j$ is a unitary.
We now name $\mathcal{L}_1=\mathcal{K}$ and set
$$w:\bigoplus_{j=1}^n\mathcal{K}\to\mathcal{H}\equiv\bigoplus_{j=1}^n\mathcal{L}_j,\;\;\;w=1_{\mathcal{L}_1}\oplus w_1\oplus w_2w_1\oplus w_3w_2w_1\oplus\dots\oplus w_{n-1}w_{n-2}\cdots w_2w_1$$
which is a unitary operator. It is obvious that $(\mathcal{H},\pi)$ is unitarily equivalent to the representation $(\bigoplus_{j=1}^n\mathcal{K},\text{ad}_w\circ\pi)$, where $\text{ad}_w(T)=w^*Tw$ for all $T\in\mathbb{B}(\mathcal{H})$.
Having another look at $(\mathcal{H},\pi)$, we see that the decomposition $\mathcal{H}=\bigoplus_{j=1}^n\mathcal{L}_j$ allows us to write an operator $T\in\mathbb{B}(\mathcal{H})$ in matrix form. More specifically,
$$T\equiv\begin{pmatrix}p_1Tp_1&p_1Tp_2&p_1Tp_3&\dots&p_1Tp_n\\ p_2Tp_1&p_2Tp_2&p_2Tp_3&\dots&p_2Tp_n\\ p_3Tp_1&p_3Tp_2&p_3Tp_3&\dots&p_3Tp_n\\ \vdots&\vdots&\vdots&\dots&\vdots\\ p_nTp_1&p_nTp_2&p_nTp_3&\dots&p_nTp_n\end{pmatrix}$$
i.e. any operator $T\in\mathbb{B}(\mathcal{H})$ is realized as a matrix $[T_{i,j}]$ where $T_{i,j}\in\mathbb{B}(\mathcal{L}_j,\mathcal{L}_i)$, $T_{i,j}=p_iTp_j$. Therefore, if $x\in M_n(A)$ is the matrix $[a_{i,j}]$, then
$$\pi(x)=[p_i\pi(x)p_j]_{i,j}$$
but now $p_i\pi(x)p_j=\pi(e_{ii}xe_{jj})=\pi(a_{ij}\otimes f_{i,j})$, where $\{f_{i,j}\}$ are the standard matrix units of $M_n(\mathbb{C})$. It is not hard to verify that, in matrix form,
$$w^*\pi([a_{i,j}])w=[p_1\pi(a_{i,j}\otimes f_{1,1})p_1]_{i,j}$$
so, if we set $\varrho:A\to\mathbb{B}(\mathcal{K})$ as
$$\varrho(a)=p_1\cdot\pi\begin{pmatrix}a&0&\dots&0\\0&0&\dots&0\\\vdots&\vdots&\dots&\vdots\\0&0&\dots&0\end{pmatrix}\cdot p_1$$
which is of course a unital representation of $A$, we then have that $w^*\pi([a_{i,j}])w=[\varrho(a_{i,j})]=\varrho^{(n)}([a_{i,j}])$, proving the claim.
 A: I think your argument is fine. Here is a slightly more compact argument that avoids having to consider the $\mathcal L_j$.
$\def\cA{\mathcal A}$ $\def\CC{\mathbb C}$ $\def\abajo{\\[0.3cm]}$ $\def\sigma{\pi}$
We have $M_n(\cA)=\cA\otimes M_n(\CC)$. Define $H_1=\sigma(I\otimes E_{11})H$, and let $W:H_1^n\to H$ be given by
$$
W(\xi_1,\ldots,\xi_k)=\sum_{k=1}^n\sigma(I\otimes E_{k1})\xi_k.
$$
Linearity is clear. It is also clear that $W$ is onto, since for any $\xi\in H$ we have
\begin{align*}
\xi&=\sigma(I\otimes I_n)\xi=\sum_{k=1}^n\sigma(I\otimes E_{kk})\xi=\sum_{k=1}^n\sigma(I\otimes E_{k1})
\sigma(I\otimes E_{1k})\xi \abajo
&=W(\sigma(I\otimes E_{11})\xi,\sigma(I\otimes E_{12})\xi,\ldots,\sigma(I\otimes E_{1n})\xi).
\end{align*}
Also
\begin{align*}
\langle W(\xi_k)_1^n,W(\eta_k)_1^n\rangle & = \sum_{k,j=1}^n\langle\sigma(I\otimes E_{k1})\xi_k,
\sigma(I\otimes E_{j1})\eta_j\rangle \abajo
&= \sum_{k,j=1}^n\langle\sigma(I\otimes E_{1j}E_{k1})\xi_k,\eta_j\rangle 
=\sum_{k=1}^n\langle\sigma(I\otimes E_{11})\xi_k,\eta_k\rangle \abajo
&=\sum_{k=1}^n \langle \xi_k,\eta_k\rangle
=\langle (\xi_k),(\eta_k)\rangle.
\end{align*}
So $W$ is a isometry and thus a unitary. Now we define $\varrho:\cA\to B(H_1)$ by
$$
\varrho(a)\eta=\sigma(a\otimes E_{11})\eta,\ \ \ \ \eta\in H_1. 
$$
It is straightforward, from the fact that $\sigma$ is a $*$-homomorphism, that so is $\varrho$. Denote by $e_1,\ldots,e_n$ the canonical basis of $\CC^n$. For any $\tilde\xi=(\xi_1,\ldots,\xi_n)\in H_1^n=H_1\otimes \CC^n$, $a\in\cA$, and $k,j\in\{1,\ldots,n\}$, \begin{align*}
\sigma(a\otimes E_{kj})W\tilde\xi &=\sum_{t=1}^n \sigma(a\otimes E_{kj})\sigma(I\otimes E_{t1})\xi_t
=\sigma(a\otimes E_{k1})\xi_j.
\end{align*}
Also, we may write $\tilde \xi=\sum_t \xi_t\otimes e_t$, and then
\begin{align*}
W\varrho^{(n)}(a\otimes E_{kj})\tilde\xi &=  W(\varrho(a)\otimes E_{kj})\sum_t \xi_t\otimes e_t
=W(\varrho(a)\xi_j\otimes e_k) \abajo
&= \sigma(I\otimes E_{k1})\varrho(a)\xi_j
=\sigma(I\otimes E_{k1})\sigma(a\otimes E_{11})\xi_j \abajo
&=\sigma(a\otimes E_{k1})\xi_j=\sigma(a\otimes E_{kj})W\tilde\xi.
\end{align*}
By linearity, we obtain $\sigma(A)W=W\varrho^{(n)}(A)$ for all $A\in M_n(\cA)$, and so $W^*\sigma(A)W=\varrho^{(n)}(A)$.
