I have a little problem with a determinant.
Let $A = (a_{ij}) \in \mathbb{R}^{(n, n)}, n \ge 4$ with
$$a_{ij} =
\begin{cases}
x \quad \mbox{for } \,i = 2, \,\, j \ge 4,\\
d \quad \mbox{for } \,i \ge j, \\
0 \quad \mbox{else.}
\end{cases}$$
So for example, if we choose $n = 5$, the matrix would look like this:
$$ A =
\begin{pmatrix}
d &0 &0 &0 &0 \\
d &d &0 &x &x \\
d &d &d &0 &0 \\
d &d &d &d &0 \\
d &d &d &d &d \\
\end{pmatrix}$$
How can I find the determinant of this matrix?
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A = T \cdot R$. Then I wanted to use $$\det(A) = \det(T \cdot R) = \det(T) \cdot \det(R).$$
to figure out the determinant. This would be something like
$$ d^n \cdot \det(R)$$
But is this approach even possible (I don't think so)? Is there any intelligent way of solving this? Thanks in advance.
Best Answer
Adding a multiple of one row to another preserves the determinant. Subtract $x/d$ of the last row from the second to get $$\begin{pmatrix} d &0 &0 &0 &0 \\ d-x &d-x &-x &0 &0 \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}$$ and then add $x/d$ of the third row to the second row to get $$\begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &0 &0 \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix}.$$ This is lower triangular, so its determinant is the product of its diagonal, which is $d^5$. This all works for the $n$ by $n$ case, so the answer in general is $d^n$.