Let ‎$‎A$ be a unital algebra and let ‎$‎n ‎\in‎ ‎N‎$‎. Then ‎$‎A‎$‎ is central if and only if ‎$‎M‎_{‎n}(A)‎$‎ is central. A nonzero unital algebra is said to be central if scalar multiples of unity are the only elements in its ‎center.‎‎‎
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Lemma ‎:‎ Let ‎‎$‎‎‎A$ be a unital algebra and let ‎‎$‎n ‎\in‎ ‎N‎$‎. Then ‎$‎A‎$‎ is central if and only if ‎‎$‎M‎_{‎‎n}(A)‎‎$‎ is central.

‎Proof:‎ If ‎‎$‎c ‎\in‎ Z(A)‎‎$‎, then the diagonal matrix with all diagonal entries equal to ‎$‎c‎$‎ lies in ‎‎$‎Z(‎M_{‎n}(A))‎‎$‎. Thus, ‎‎$‎‎A‎$‎ is central if ‎‎$‎M_{‎n}(A)‎$‎ is. Now let ‎‎$‎A‎$‎ be central. Pick ‎‎$‎C ‎\in‎ Z(‎M_{‎n}(A))‎‎$‎. Since ‎‎$‎C‎$‎ commutes with every ‎‎$‎E_{‎kl} , k ‎\neq‎l ‎‎‎‎$‎, one easily shows that ‎‎$‎C‎$‎‎ is a diagonal
matrix with all diagonal entries equal. As ‎‎$‎C‎$‎‎ also commutes with every ‎‎$‎aE_{‎11}, a ‎\in ‎A‎$‎,we conclude that this diagonal entry lies in the center of ‎‎$‎A‎$‎, and hence it is a scalar.Thus $‎M‎_{‎‎n}(A)‎$‎ is ‎central.‎‎
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1:Why ‎c‎an ‎we ‎say ‎"‎‎If ‎‎$‎c ‎\in‎ Z(A)‎‎$‎, then the diagonal matrix with all diagonal entries equal to ‎$‎c‎$‎ lies in ‎‎$‎Z(‎M_{‎n}(A))‎‎$ ‎and ‎‎$‎‎A‎$‎ is ‎central‎ ‎"?‎‎ 
2:Why ‎is ‎‎$‎C‎$‎ a diagonal matrix with all diagonal entries ‎equal , ‎‎‎and ‎why dose ‎the ‎diagonal entry ‎lie   in the center of ‎‎$‎A$‎‎‎?‎

 A: These follow from direct and simple computations. You should just do them.
1) If $M$ is a matrix in $M_n(A)$, and $C$ represents a matrix with element $c$ down the diagonal, then 
$$
[CM]_{ij}=cM_{ij}=M_{ij}c=[MC]_{ij}
$$
2) If $C$ has to commute with every $M$, then in particular it commutes with the matrix with a $a$ in its upper left corner and zeros elsewhere, for every $a\in A$. Then $CM$ and $MC$ are equal, so their upper left hand corners, $ac$ and $ca$ are equal. That would imply that $c$ is central.
A: It may be clearer to claim that

The map of rings $D:A\to M_n(A)$ that sends $a$ to $D(a) = \text{diag}(a,\ldots,a)$ restricts to an isomorphism of rings $D: Z(A) \to Z(M_n(A))$. That is, the center of $Z(M_n(A))$ consists of diagonal matrices of the form $\text{diag}(a,\ldots,a)$ with $a\in Z(A)$. In particular $A$ is central if and only if $M_n(A)$ is central. 

If $a\in Z(A)$, then a direct computation shows that $D(a)$ commutes with every matrix in $M_n(A)$. 
Conversely, let $M$ be a matrix in $M_n(A)$ that commutes with every matrix, and let us write $MN-NM = [M,N]$, the commutator of $M$ and $N$. Saying that $M$ and $N$ commute means exactly this commutator is $0$.
In particular, if $M$ commutes with every matrix, it commutes with the matrices $e_{ij}$, $i\neq j$. Recall from linear algebra that $e_{ij}M$ is the matrix that has row $j$ of $M$ in row $i$ and is zero elsewhere, and that $Me_{ij}$ is the matrix that has column $i$ of $M$ in column $j$ is zero elsewhere.
It follows that $[M,e_{ij}]$ is the following matrix
\begin{pmatrix}
0 &\cdots & 0 &M_{1i} & 0 & \cdots & 0 \\
0 &\cdots & 0 &M_{2i} & 0 & \cdots & 0 \\
\vdots & \ddots &\vdots &\vdots&\vdots &\ddots&\vdots\\
-M_{j1} & -M_{j2} &\cdots &0&\cdots&\cdots&-M_{jn} \\
\vdots & \ddots &\vdots &\vdots&\vdots &\ddots&\vdots\\
0 &\cdots & 0 &M_{ni} & 0 & \cdots & 0 
\end{pmatrix}
Doing this for every $i\neq j$ proves that $M$ has zeros outside the diagonal,
so $M$ is a diagonal matrix, say $\text{diag}(a_1,\ldots,a_n)$. Now taking matrices of the form $D(a)$ and computing that $$[M,D(a)] = \text{diag}([a_1,a],\ldots,[a_n,a])$$
shows that $a_i\in Z(A)$ for each $i\in\{1,\ldots,n\}$, thus completing the proof. 
