Null Space of A Matrix
The column and row rank are always the same. In the case, by eliminating the two rows that are multiplier of the others, we get the rank of the matrix is $3$. The null space has a dimension of $5-3=2$.
Probability
Gaussian Noise to Regularize Parameters
Adding noise to the predictors is equivalent to conduct ridge regression: https://stats.stackexchange.com/questions/548756/how-is-adding-noise-to-training-data-equivalent-to-regularization
Linear Regression - Heteroskedastic
Average Time Between Customers to Supervisor
Trace of Matrix Exponential
Reference: https://math.stackexchange.com/questions/1821508/trace-of-matrix-exponential-closed-form-expression
Conclusion is, $$ \operatorname{Tr}(\exp (A))=\sum_{\lambda \in \Lambda(A)} m_{\lambda} e^{\lambda} $$ where $\Lambda(A)$ is the set of distinct generalized eigenvalues of $A$, and $m_{\lambda}$ is the multiplicity corresponding to $\lambda \in \Lambda(A)$.
For this problem, we can find for matrix $A$, the two eigenvalues are $1/2$ and $1$.So trace of $e^A$ is $T =e^{0.5}+e^1=4.3670$. Then $3T=13.101$.