Determinant of a 5 × 5 matrix

Question

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.

Answer 1

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 $d^n$.

Answer 2

Develop your matrix wrt the first row and get

$$|A|=d\begin{vmatrix}d&0&x&x\\d&d&0&0\\d&d&d&0\\d&d&d&d\end{vmatrix}$$

Develop again wrt the first row but observe that when your pivot points are the $\;x$'s you get determinant zero as there are two identical rows in each case, so we get

$$d^2\begin{vmatrix}d&0&0\\d&d&0\\d&d&d\end{vmatrix}=d^5$$

Try now some inductive argument based on this.

Answer 3

My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A=T⋅R$.

That's very much a way to do it. The technique is called LU Decomposition. It produces a lower and upper triangular matrix, allowing trivial determinate calculations. For this reason, you actually only need to find the diagonal elements to get your determinant.

In this case,

$$ A = \begin{pmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{pmatrix} = \begin{pmatrix} 1 &0 &0 &0 &0 \\ 1 &1 &0 &0 &0 \\ 1 &1 &1 &0 &0 \\ 1 &1 &1 &1 &0 \\ 1 &1 &1 &1 &1 \\ \end{pmatrix} \cdot \begin{pmatrix} d &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ 0 &0 &d &-x &-x \\ 0 &0 &0 &d &0 \\ 0 &0 &0 &0 &d \\ \end{pmatrix}$$

So

$$ \det A = \det L \cdot \det U = 1^5 \cdot d^5 $$

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The first few steps of the method used here (takes longer to texify than to do) are: $$ \begin{eqnarray} \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ d &d &0 &x &x \\ d &d &d &0 &0 \\ d &d &d &d &0 \\ d &d &d &d &d \\ \end{smallmatrix}\right] &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ 0 &d &d &0 &0 \\ 0 &d &d &d &0 \\ 0 &d &d &d &d \\ \end{smallmatrix}\right] \\ &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &d &0 &x &x \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 &0 &0 &0 &0 \\ 0 &0 &0 &0 &0 \\ 0 &0 &d &-x &-x \\ 0 &0 &d &d-x &-x \\ 0 &0 &d &d-x &d-x \\ \end{smallmatrix}\right] \\ &&= \left[\begin{smallmatrix} d &0 &0 &0 &0 \\ 0 &d &0 &x &x \\ \end{smallmatrix}\right] \left[\begin{smallmatrix} 1 & 0\\ 1 & 1\\ 1 & 1\\ 1 & 1\\ 1 & 1\\ \end{smallmatrix}\right] + \cdots \end{eqnarray} $$