How do I write this multiplication of matrices down correctly: $(A \cdot B)^{2} = (A \cdot B) \cdot (A \cdot B)$?

Question

Two real matrices are given:

$$A=\begin{pmatrix} 2 & 5 & -3\\ 0 & -1 & 6 \end{pmatrix} \text{ and } B=\begin{pmatrix} 3 & 1\\ 4 & -2\\ 0 & 6 \end{pmatrix}$$

Now I want show that $(A \cdot B)^{2}= (A \cdot B) \cdot (A \cdot B)$

The problem is I don't know if we are allowed to say that both sides are the same. That means I have to do it on both sides I think :(

Or what do you think?

Assuming I have to do it on both sides, how would the left side look then, correctly written as matrices?

Would this notation be correct? $$(A \cdot B)^{2} = \begin{pmatrix} 2 \cdot 3 + 5 \cdot 4 -3 \cdot 0 & 2 \cdot 1+5(-2) -3 \cdot 6 \\ 0 \cdot 3 -1 \cdot 4+6 \cdot 0 & 0 \cdot 1-1(-2)+6 \cdot 6 \end{pmatrix}^{2}= \begin{pmatrix} 26 & -26\\ -4 & 38 \end{pmatrix}^{2}= \begin{pmatrix} 26 & -26\\ -4 & 38 \end{pmatrix} \cdot \begin{pmatrix} 26 & -26\\ -4 & 38 \end{pmatrix}= \begin{pmatrix} 26 \cdot 26 -26(-4) & 26(-26)-26 \cdot 38\\ -4 \cdot 26+38 \cdot (-4) & -4 \cdot (-26)+38 \cdot 38 \end{pmatrix}= \begin{pmatrix} 780 & -1664\\ -256 & 1548 \end{pmatrix}$$

Can I write it like that and do I have to calculate the other side really too?

Answer 1

One does not need to do any calculations but noting that by the definition of "squaring" square matrices (in particular $2\times 2$ matrices): $$ C^2:=C\cdot C\tag{1} $$ Now let $C=A\cdot B$. Then (1) implies that $$ (A\cdot B)^2=(A\cdot B)\cdot(A\cdot B) $$

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Remark. In linear algebra, one usually doesn't write the dot $\cdot$ in matrix multiplication $A\cdot B$ since the dot is preserved for "dot product" of vectors.

Answer 2

Isn't "the right side" exactly what you did? That is, you multiplied A and B to get AB and them multiplied them together to get (AB)^2. That is precisely (AB)(AB). You could also write that as ABAB which you would do by first multiplying AB together, then multiply B times that: B(AB), then multiply A by that A(BAB)= ABAB.