Evaluation of determinant

Evaluate the determinant of the $n \times n$ matrix:

$$A=\left({\begin{array}{ccccccc}
1+a^2 & a & 0 & 0 & \cdots & 0 & 0\\
a & 1+a^2 & a & 0 & \cdots & 0 & 0\\
0 & a & 1+a^2 & a & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 &  0 & 0 & \cdots &   a & 0\\
0 & 0 &  0 & 0 & \cdots &  1+a^2 & a\\
0 & 0 &  0 & 0 & \cdots  &   a & 1+a^2
\end{array}}\right)\,,\quad a\in\mathbb{R}$$

Solution



We expand the determinant with respect to the elements of the first row of $A$ and we have:

$$\left|A_{n}\right|=(-1)^{1+1}(1+a^2)\left|A_{n-1}\right|+(-1)^{2+1}a\left|B\right|=(1+a^2)\left|A_{n-1}\right|-a\left|B\right|$$

where $B$ is the following matrix:

$$B=\begin{pmatrix}
 a & a & 0 & \cdots & 0 & 0 & 0\\
0 & 1+a^2 & a & \cdots & 0 & 0& 0\\
\vdots & \vdots & \vdots &\vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1+a^2& a & 0\\
0 & 0 & 0 & \cdots & a & 1+a^2 & a\\
 0 & 0 & 0 & \cdots & 0 & a & 1+a^2
 \end{pmatrix}$$

For the determinant of $B$ we have:

$$\left|B\right|=a\left|A_{n-2}\right|$$

Hence we get the recurrence relation:

$$\left|A_{n}\right|=(1+a^2)\left|A_{n-1}\right|-a^2\left|A_{n-2}\right|  \tag{1}$$

Using the recurrence relation we have successively:

$$\begin{array}{rl}
|{A_{n}}|  \hspace{-0.2cm}& = ({1+a^2})\,|{A_{n-1}}|-a^2\, |{A_{n-2}}|\\
           & \stackrel{(1)}{=\!=}  ({1+a^2})\,\bigl({({1+a^2})\,|{A_{n-2}}|-a^2\, |{A_{n-3}}|}\bigr)-a^2\, |{A_{n-2}}|\\
           & =  ({1+a^2+a^4})\,|{A_{n-2}}|-a^2\,({1+a^2})\,|{A_{n-3}}|\\
           & \stackrel{(1)}{=\!=}  ({1+a^2+a^4})\,\bigl({({1+a^2})\,|{A_{n-3}}|-a^2\, |{A_{n-4}}|}\bigr)-a^2\,({1+a^2})\,|{A_{n-3}}|\\
           & =  ({1+a^2+a^4+a^6})\,|{A_{n-3}}|-a^2\,({1+a^2+a^4})\,|{A_{n-4}}|\\
           & =  ({1+a^2+a^{2\cdot2}+a^{2\cdot3}})\,|{A_{n-3}}|-a^2\,({1+a^2+a^{2\cdot2}})\,|{A_{n-4}}|\\
           & ........................................................... \\
           & =\bigl({1+a^2+a^{2\cdot2}+\ldots+a^{2(n-2)}}\bigr)\,|{A_{2}}|-a^2\,\bigl({1+a^2+\ldots+a^{2(n-3)}}\bigr)\,|{A_{1}}|\\
           & =\bigl({1+a^2+a^{2\cdot2}+\ldots+a^{2(n-2)}}\bigr)\,({1+a^2+a^4})-a^2\,\bigl({1+a^2+\ldots+a^{2(n-3)}}\bigr)\,({1+a^2})\\
           & =1+a^2+a^{2\cdot2}+\ldots+a^{2(n-2)}+a^{2(n-1)}+a^{2n}
\end{array}$$

and the exercise is complete.