Given A = [[9,1,1],[1,2,1],[1,18,1]] and B = the 3x2 matrix of all ones, compute AB, and then decide whether A^T B^T is defined.

Step-by-step solution

1. 1

   Check the shapes before multiplying. $A$ is $3\times 3$ and $B$ is $3\times 2$. The inner dimensions match ($3=3$), so $AB$ exists and has the outer dimensions, $3\times 2$.

2. 2

   Exploit the structure of B. Every column of $B$ is $(1,1,1)^{t}$, so each entry of $AB$ is just the sum of a row of $A$, and both columns of $AB$ will be identical.

3. 3

   Compute the three row sums.

   $9+1+1=11,\qquad 1+2+1=4,\qquad 1+18+1=20$

   Hence

   $AB=\begin{pmatrix}11&11\\4&4\\20&20\end{pmatrix}$

4. 4

   (b) Check the shapes for AᵗBᵗ. Transposing swaps dimensions: $A^{t}$ is $3\times 3$ and $B^{t}$ is $2\times 3$. The product $A^{t}B^{t}$ would need the inner dimensions to agree, i.e. $3$ (columns of $A^t$) $=2$ (rows of $B^t$) — and they do not. So

   $A^{t}B^{t}\ \textbf{is undefined.}$

5. 5

   State the identity that does hold. Transposing a product reverses the order:

   $(AB)^{t}=B^{t}A^{t}$

   Here $B^{t}$ is $2\times 3$ and $A^{t}$ is $3\times 3$, so $B^{t}A^{t}$ is $2\times 3$ — exactly the shape of $(AB)^{t}$.

6. 6

   Verify the identity numerically.

   $B^{t}A^{t}=\begin{pmatrix}11&4&20\\11&4&20\end{pmatrix}=(AB)^{t}$

   which is indeed the transpose of the $AB$ found in part (a). This confirms both the arithmetic and the order-reversal rule.

   $\boxed{AB=\begin{pmatrix}11&11\\4&4\\20&20\end{pmatrix};\quad A^{t}B^{t}\text{ undefined};\quad (AB)^{t}=B^{t}A^{t}}$