I was given this problem in my Linear Algebra 1 homework:
Calculate the determinant of a general matrix of order $n\times n$ that looks like this:
$$
\left(
\begin{matrix}
a+b & b & \cdots & \cdots & b \\
a & a+b & b & \cdots & \vdots \\
\vdots & a & \ddots & \ddots & \vdots \\
a & \ddots & \ddots & \ddots & b \\
a & a & \cdots & a & a+b \\
\end{matrix}
\right)
$$
There is a given hint which is to find a recurrence relation and solve it.
What I did so far:
I simplified the matrix using the elementary operation of row addition that doesn't change the determinant to this:
$$
\left(
\begin{matrix}
a+b & b & b & \cdots & \cdots & b \\
-b & a & 0 & 0 & \cdots & \vdots \\
0 & -b & a & 0 & \cdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & \ddots & 0 \\
0 & 0 & \cdots & 0 & -b & a \\
\end{matrix}
\right)
$$
And then by using the definition of a determinant according to the last row I developed it into this:
$$
b \cdot
\left|
\begin{matrix}
a+b & b & b & \cdots & \cdots & b \\
-b & a & 0 & 0 & \cdots & \vdots \\
0 & -b & a & 0 & \cdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & \ddots & 0 \\
0 & 0 & \cdots & 0 & -b & 0 \\
\end{matrix}
\right|
+ a \cdot
\left|
\begin{matrix}
a+b & b & b & \cdots & \cdots & b \\
-b & a & 0 & 0 & \cdots & \vdots \\
0 & -b & a & 0 & \cdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & \ddots & 0 \\
0 & 0 & \cdots & 0 & -b & a \\
\end{matrix}
\right|
$$
I figured out that the matrix that is being multiplied by a is just the same matrix of order (n-1)x(n-1) so the recurrence relation is $a_n = b\cdot |B| + a\cdot a_{n-1}$ where $B$ is the matrix on the left
I'm stuck on this last step of trying to figure out the value of $|B|$
Can anyone help?
Best Answer
You are close to the solution of your problem. The second $n\times n$ determinant with lots of zeros can be expanded in more than one way. Perhaps the easiest is to expand by the first column. The result is $(a+b)$ times the determinant of $A_{n-1}$, an $(n-1)\times(n-1)$ lower triangular matrix with all $a\,$s on the diagonal plus $b$ times the determinant of $B_{n-1}$, an $(n-1)\times(n-1)$ matrix which is the same as the lower triangular matrix except the first row is all $b\,$s.
Let $u_n$ be the determinant of the original $n\times n$ matrix. The determinant of $A_{n-1}$ is easily $a^{n-1}$. Let $v_{n-1}$ be the determinant of $B_{n-1}$. Using expansion by the first column get the recursion $v_n = ba^{n-1} +bv_{n-1}$ with initial value $v_1 = b.$ The recursion for $u_n$ is $u_n = (a+b)a^{n-1} +bv_{n-1}$ with initial value $u_1 = a+b.$ You can verify that $u_n = v_n + a^n$ and $v_n = b u_{n-1}$ for all $n>0.$