The dimensions of A and B are as follows: A= 3x3, B= 3x1

When we mulitply two matrices, we need to keep in mind their dimensions (in this case 3x3 and 3x1).

The two inner numbers need to be the same. Otherwise, we cannot multiply them. The product's dimensions will be the two outer numbers: 3x1. 

The dimensions of both A and B are 2x2. Therefore, the matrix that results from their product will have the same dimensions. 

\(\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f \\ g & h \end{bmatrix}=\begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}\)

Thus plugging in our values for this particular problem we get the following:

The dimensions of A and B are as follows: A=1x3, B= 3x1. 

Because the two inner numbers are the same, we can find the product. 

The two outer numbers will tell us the dimensions of the product: 1x1. 

\(\displaystyle \begin{bmatrix} a & b & c \end{bmatrix} \cdot \begin{bmatrix} d \\ e\\ f\end{bmatrix}= \begin{bmatrix} ad + be +cf \end{bmatrix}\)

Therefore, plugging in our values for this problem we get the following:

The dimensions of A and B are as follows: A= 3x1, B= 2x3

In order to be able to multiply matrices, the inner numbers need to be the same. In this case, they are 1 and 2. As such, we cannot find their product.

The answer is No Solution. 

To have the above equality we need to have \(\displaystyle x^{2}=x\) and \(\displaystyle x=3\).

\(\displaystyle x^2=x\) means that \(\displaystyle x=0\), or \(\displaystyle x=1\). Trying all different values of \(\displaystyle x\), we see that no \(\displaystyle x\) can satisfy both matrices.

Therefore there is no \(\displaystyle x\) that satisfies the above equality.

We will use an induction proof to show this result.

We first note the above result holds for n=1. This means \(\displaystyle A^{1}=\begin{bmatrix} 1 &1 \\ 1 &1 \end{bmatrix}\)

We suppose that  \(\displaystyle A^n=\begin{bmatrix} 2^{n-1}& 2^{n-1} \\ 2^{n-1}& 2^{n-1} \end{bmatrix}\) and we need to show that:\(\displaystyle A^{n+1}=\begin{bmatrix} 2^{n} &2^{n} \\2^{n}&2^{n} \end{bmatrix}\)

By definition \(\displaystyle A^{n+1}=A^{n}A\). By inductive hypothesis, we have:

\(\displaystyle A^{n}=\begin{bmatrix} 2^{n-1}&2^{n-1}\\ 2^{n-1}&2^{n-1}\end{bmatrix}\)

Therefore, \(\displaystyle A^{n+1}=\begin{bmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} &2^{n} \end{bmatrix}\begin{bmatrix} 1 & 1\\ 1&1 \end{bmatrix}\)

\(\displaystyle A^{n+1}=\begin{bmatrix} 2^{n} &2^{n} \\ 2^{n} &2^{n} \end{bmatrix}\)

This shows that the result is true for n+1. By the principle of mathematical induction we have the result.

Note that:

 \(\displaystyle \begin{bmatrix} 1 & 0 &0 &0 &0 \\ 1 & 0 &0 &0 &0 \\1 & 0&0 &0 &0 \\1 &0 &0 &0 &0 \\1 &0 &0 &0 &0 \end{bmatrix} \begin{bmatrix} 1 & 0 &0 &0 &0 \\ 1 & 0 &0 &0 &0 \\1 & 0&0 &0 &0 \\1 &0 &0 &0 &0 \\1 &0 &0 &0 &0 \end{bmatrix}=\)    \(\displaystyle \begin{bmatrix} 1 & 0 &0 &0 &0 \\ 1 & 0 &0 &0 &0 \\1 & 0&0 &0 &0 \\1 &0 &0 &0 &0 \\1 &0 &0 &0 &0 \end{bmatrix}\)

Since \(\displaystyle A^{4}=A^{2}A^{2}=AA=A^{2}=A\).

This means that \(\displaystyle A^{4}=A\)

We know that to be able to have the product of the 2 matrices, the size of the column of A must equal the size of the row of B. This gives :

\(\displaystyle n^{2}=n+2\).

Solving for n, we find

\(\displaystyle n^2-n-2=0\)

\(\displaystyle (n-2)(n+1)=0\)

\(\displaystyle n=2, n=-1\)

Since n is a natural number \(\displaystyle n=2\) is  the only possible solution.

Note that every entry of the product matrix is the sum of \(\displaystyle 2+2+...2\) (\(\displaystyle n\) times) .

\(\displaystyle AB=\begin{bmatrix} 1 &1 &\cdots &1 \\ 1 &1 &\cdots &1 \\\vdots &\vdots &\vdots &\vdots \\ 1 &1 &\cdots &1 \end{bmatrix} \times \begin{bmatrix} 2 &2 &\cdots &2 \\ 2 &2 &\cdots &2 \\\vdots &\vdots &\vdots &\vdots \\ 2&2 &\cdots &2 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2 \\ 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2\\\vdots& \vdots\end{bmatrix}=\begin{bmatrix} 2n & 2n &2n ... \\2n & 2n &2n ... \\\vdots &\vdots\end{bmatrix}\)

This gives \(\displaystyle 2n\) as every entry of the product of the two matrices.

\(\displaystyle AB=\begin{bmatrix} 1&1&\cdots &1\\ 0 &0&\cdots &0\\\vdots &\vdots &\vdots &\vdots \\ 0 &0 &0 &0 \end{bmatrix} \times\begin{bmatrix} 1&0 &\cdots &0\\ 1 &0 &\cdots &0 \\\vdots &\vdots &\vdots &\vdots \\ 1&0 &0 &0 \end{bmatrix}\)

\(\displaystyle =\begin{bmatrix}1(1) +1(1)+...1(1) & 1(0)+0(0)+..0(0)\\ 0(1)+0(0)+...0(0) \\\vdots& \vdots \end{bmatrix}=\begin{bmatrix} 1(n) &0 &...\\0&0&...\\\vdots&\vdots\end{bmatrix}\)

Note that when we multiply the first row by the first colum we get: \(\displaystyle 1+1...1\) (\(\displaystyle n\) times), this gives the value of \(\displaystyle n\).

All other rows are zeros, and therefore we have zeros in the other entries.