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Consider $$A$$ is a square matrix of order $$n×n$$ and $$I$$ is the unit matrix of the same order then $$AI = IA = A$$.

Here, the $$2 × 2$$ order of $$I$$ matrix is $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Then, the $$3 × 2$$ order of $$I$$ matrix will be $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

Now let's prove $$AI = IA = A$$ using $$2 × 2$$ order matrix.
Example:
If $$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ then we have:

$\begin{array}{l}\mathit{AI}=\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left[\begin{array}{ll}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ \\ \mathit{AI}=\left[\begin{array}{ll}\left(1×1\right)+\left(2×0\right)& \left(1×0\right)+\left(2×1\right)\\ \left(3×1\right)+\left(4×0\right)& \left(3×0\right)+\left(4×1\right)\end{array}\right]\phantom{\rule{0.147em}{0ex}}\end{array}$

$$AI = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$……(1)

Let's find $$IA$$ matrix.

$\begin{array}{l}\mathit{IA}=\phantom{\rule{0.147em}{0ex}}\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\phantom{\rule{0.147em}{0ex}}\left[\begin{array}{ll}1& 2\\ 3& 4\end{array}\right]\\ \\ \mathit{IA}=\left[\begin{array}{ll}\left(1×1\right)+\left(0×3\right)& \left(1×2\right)+\left(0×4\right)\\ \left(0×1\right)+\left(1×3\right)& \left(0×2\right)+\left(1×4\right)\end{array}\right]\phantom{\rule{0.147em}{0ex}}\end{array}$

$$IA = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$……(2)

From $$(1)$$ and $$(2)$$ we get that $$(1) = (2) = A$$.

Hence, $$AI = IA = A$$ proved.