I have this question here:
Suppose that $A$ is an $(n,n)$-matrices, $n \geq 3$, with $\det(A)=5$ and whose $(2,3)$-minor $M_{2,3}(A)$ equals $12$. Let $S_{2,3}$ be the $(n,n)$-matrix all whose entries are $0$, except the entry
in position $(2,3)$ equals $2$. With $B:=(A+S_{2,3})$, find $\det(B)$.
The problem is that I am not sure how to go about this. I know for a fact that if $C=A+B$ in general, then $\det(C) \neq \det(A) + \det(B)$ so thus:
$\det(B) = \det(A+S_{2,3})$
I tried a sample matrix:
If $A=\begin{bmatrix}4 & 1 & 2 \\ 3 & 1 & 1 \\ 4 & 4 & 1\end{bmatrix}$
The determinant of this matrix is $5$ and the minor $M_{2,3}(A)$ is $A=\begin{bmatrix}4 & 1\\ 4 & 4\end{bmatrix}$. The determinant of this minor is $12$.
$S_{2,3} = A=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{bmatrix} $
Thus, $A + S_{2,3} = B =\begin{bmatrix}4 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 4 & 1\end{bmatrix}$. The determiant of this matrix gives $-19$. However, is there a better way of doing this in general? I don't think using specific matrices is the best idea.
Best Answer
A faster way then using the Laplace expansion (cofacter expansion), would be first using row operations and then take the determinant in REF. A determinant of a triangular matrix is just the product of the diagonal. Hope this may help.