The question is given below:
My question is:
I have calculated the determinant for $n=3$ and $n=4$ and I can guess that we have the following recurrence relation $$\det (A_{n}) = \det (A_{n-1}) - \det (A_{n-2})$$, but the question said show that, so how can I prove this recurrence relation?
In general: My question is how to prove this recurrence relation?
EDIT:
Also, I think the statement of the question is wrong as it is also $-1$ in case of $n =4$.
On
Take a matrix of an unspecified dimension $n$ and develop the determinant along the first row.
It demonstrates, as in the other answer, that your recurrence is exact.
However the starting conditions are $D_1 = 1,\; D_2 = 0$, so the complete recurrence is $$ D_{\,n} = D_{\,n - 1} - D_{\,n - 2} + \left[ {n = 1} \right] - \left[ {n = 2} \right] $$ where $[P]$ denotes the Iverson bracket $$ \left[ P \right] = \left\{ {\begin{array}{*{20}c} 1 & {P = TRUE} \\ 0 & {P = FALSE} \\ \end{array} } \right. $$
The result so is $$ D_{\,n} = \left\{ {\matrix{ {\left( { - 1} \right)^{\,k} } & {n = 3k + 1} \cr 0 & {n = 3k + 2} \cr { - \left( { - 1} \right)^{\,k} } & {n = 3k + 3} \cr } } \right.\quad \left| {\;0 \le k} \right. $$ or $$ D_{\,n + 1} = \left( { - 1} \right)^{\,\left\lfloor {n/3} \right\rfloor } \left\{ {\matrix{ 1 & {n \equiv 0\;\;\left( {\bmod 3} \right)} \cr 0 & {n \equiv 1\;\;\left( {\bmod 3} \right)} \cr { - 1} & {n \equiv 2\;\;\left( {\bmod 3} \right)} \cr } } \right.\quad \left| {\;0 \le n} \right. $$
To prove the relationship you can simply multiply out the determinant.
For $n\ge3$, $\det (A_{n}) = a_{11}\det (A_{n-1})-a_{12}a_{21}\det (A_{n-2})$
Therefore
$\det (A_{n}) = \det (A_{n-1}) - \det (A_{n-2})$.
$\det (A_{n})$ is then $0$ if $n=3k+2$ and is $(-1)^{k}$ if $n=3k$ or $n=3k+1.$
Proof using induction
Assume the result to be true for all values smaller than $n$. We know the result is true for $n=1,2,3$.
There are six cases. First suppose $n=6k+1$.
$\det (A_{6k+1}) = \det (A_{6k}) - \det (A_{6k-1})=1-0=1$.
Next suppose $n=6k+2$.
$\det (A_{6k+2}) = \det (A_{6k+1}) - \det (A_{6k})=$ etc.
You obtain 6 results all of which will be in agreement with our stated result. This might look involved but all we are doing is proving that the sequence of determinants is 1,0,-1,-1,0,1,1,0,-1,-1,0, 1, 1, ...