Given $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm B \in \mathbb R^{m \times p}$, we form a matrix equation in $\mathrm X \in \mathbb R^{n \times p}$
$$\mathrm A \mathrm X = \mathrm B$$
If we want to find the least-squares solution, then we minimize $\| \mathrm A \mathrm X - \mathrm B \|_F$. However, if we do not want the norm of $\mathrm X$ to become too large, then we minimize the following objective function
$$\begin{array}{rl} \| \mathrm A \mathrm X - \mathrm B \|_F^2 + \lambda \|\mathrm X\|_F^2 &= \mbox{tr} ((\mathrm A \mathrm X - \mathrm B)^T (\mathrm A \mathrm X - \mathrm B)) + \lambda \,\mbox{tr} (\mathrm X^T \mathrm X)\\ &= \mbox{tr} (\mathrm X^T (\mathrm A^T \mathrm A + \lambda \mathrm I_n) \mathrm X - \mathrm B^T \mathrm A \mathrm X - \mathrm X^T \mathrm A^T \mathrm B + \mathrm B^T \mathrm B)\end{array}$$
where $\lambda > 0$. Differentiating with respect to $\mathrm X$, we obtain
$$2 (\mathrm A^T \mathrm A + \lambda \mathrm I_n) \mathrm X - 2 \mathrm A^T \mathrm B$$
Finding where the derivative vanishes, we obtain the matrix equation
$$(\mathrm A^T \mathrm A + \lambda \mathrm I_n) \mathrm X = \mathrm A^T \mathrm B$$
If $\lambda > 0$, then $\mathrm A^T \mathrm A + \lambda \mathrm I_n$ is always invertible. Hence, the unique minimizer is
$$\hat{\mathrm X} := (\mathrm A^T \mathrm A + \lambda \mathrm I_n)^{-1} \mathrm A^T \mathrm B$$