Is there a general method to find a 3x3, or 2x2 matrices, given the determinant? I want to do a project with my students when we start to study Systems of Equations. It would be interesting if the determinant is a prime number and then work backwards.
For example, I got this by playing around: if I let
$$A = \begin{bmatrix} 10 & -17\\ 31 & 2 \end{bmatrix}$$
then the $\det(A)=547$, a prime number. Not sure if this would be a good mini project or not. Thanks for any help.
On
Recall if you let $A$ be a $n \times n$ matrix in lower (or upper) triangular form, then $$\det(A) = \prod_{i = 1}^n a_{ii}.$$ Using this, you can select your entries along the diagonal such that their product equals whatever value you desire and then you may fill in the remaining entries of the matrix however you desire.
Start with the diagonalized form $A=PDP^{-1}$ where $D=\mathrm{diag}([\lambda_1, \lambda_2,\cdots,\lambda_n])$. You can then choose the diagonal terms of $D$ such that $\det(A)=\lambda_1 \lambda_2\cdots\lambda_n$. Finally you can create a similar matrix by choosing any invertable $P$.