Let ${\bf A} := \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}$ and let ${\bf A}^\top$ be its transpose. Find the determinant of the matrix ${\bf B} := {\bf A} {\bf A}^\top + {\bf I}_4$, where ${\bf I}_4$ is the $4 \times 4$ identity matrix.
This seems impossible to me since ${\bf A} {\bf A}^\top$ in this case would be $4$ which you can't add to ${\bf I}_4$ since they have different dimensions. Perhaps I'm missing something.
On
No, you're correct. In general, the product $AB$ is defined for $A \in \mathbb{F}^{m \times n}$ and $B \in \mathbb{F}^{n \times r}$, and gives a matrix $AB \in \mathbb{F}^{m \times r}$.
As written, you have $A \in \mathbb{R}^{1 \times 4} \implies A^{\sf{T}} \in \mathbb{R}^{4 \times 1}$, so $AA^{\sf{T}} \in \mathbb{R}^{1 \times 1}$ (or, for all intents and purposes, just $\mathbb{R}$), so you certainly can't add a $4 \times 4$ matrix to it.
The most likely conclusion is that it was intended that $A$ be a column vector, i.e. $(1,1,1,1)^{\sf{T}}$, and so $AA^{\sf{T}}$ would be a $4 \times 4$ matrix.
A matrix $A$ of size $m_1 \times n_1$ can be multiplied by a matrix $B$ of size $m_2 \times n_2$ to produce a product $AB$ only if $n_1 = m_2$ (the number of columns of $A$ is equal to the number of rows of $B$).
The resulting product then has size $m_1 \times n_2$. So here, if $A$ is a column vector with size $4 \times 1$, then $A^T$ is a row vector with size $1 \times 4$ and the resulting product $AA^T$ has size $4 \times 4$.
As you say, if $A$ is a row vector with size $1 \times 4$, then the product $AA^T$ produces a matrix of size $1 \times 1$, or a scalar.
Perhaps the question is incorrect.