# How to solve these general $n \times n$ determinants?

How would you solve these two general determinants?

$$\begin{vmatrix} 2 & 1 & 0 & \cdots & 0 & 0 \\ 1 & 2 & 1 & \cdots & 0 & 0\\ 0 & 1 & 2 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 2 & 1\\ 0 & 0 & 0 & \cdots & 1 & 2\\ \end{vmatrix}$$

$$\begin{vmatrix} 3 & 2 & 2 & \cdots & 2 \\ 2 & 3 & 2 & \cdots & 2 \\ 2 & 2 & 3 & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & 3 \\ \end{vmatrix}$$

And if there are any tips for counting determinants of this type, please let me know :-)

• use cofactor expansion and check which minor matrices have linear dependant rows/colums. en.wikipedia.org/wiki/… – Max Jan 1 '17 at 12:20
• The first one is solved here: math.stackexchange.com/questions/266998/… – Martin R Jan 1 '17 at 12:20
• my comment above should have been an answer btw^^ – Max Jan 1 '17 at 12:29
• Have you attempted something ? – Jean Marie Jan 1 '17 at 13:12

For the second one, in general, if $$\Delta_n=\begin{vmatrix} a & b & b & \ldots & b\\ b & a & b & \ldots & b \\ b & b & a & \ldots & b \\ \vdots&&&&\vdots\\ b & b & b & \ldots & a \end{vmatrix}.$$ First step
$R_i\to R_i-R_n\;(i\ne n)$ implies $$\Delta_n= \begin{vmatrix} a & b & b & \ldots & b\\ b -a& a-b & 0 & \ldots & 0 \\ b-a & 0 & a-b & \ldots & 0 \\ \vdots&&&&\vdots\\ b -a& 0 & 0 & \ldots & a -b\end{vmatrix}.$$ Second step
$C_1\to C_1+C_2+\cdots +C_n$ implies $$\Delta_n= \begin{vmatrix} a +(n-1)b& b & b & \ldots & b\\ 0 & a-b & 0 & \ldots & 0 \\ 0 & 0 & a-b & \ldots & 0 \\ \vdots&&&&\vdots\\ 0 & 0 & 0 & \ldots & a -b\end{vmatrix}=[a+(n-1)b](a-b)^{n-1}.$$

• What operation are you doing in the second step? I understand the result of it but I am not sure with the process. – Marek Otypka Jan 1 '17 at 13:32
• To the firs column add the other ones – Fernando Revilla Jan 1 '17 at 16:48
• Yes, I finally see it, thank you! :-) – Marek Otypka Jan 1 '17 at 22:14

Hint for the first determinant:

The first determinant is a tridiagonal determinant $A_n=\det(a_{i,j})_{1\le i,\mkern2mu j\le n}$, where $a_{i,\mkern2mu j}=0\;$ if $\lvert i-j\rvert>1$. You can easily prove, with Laplace's expansion along the last row, the linear recurrence relation: $$A_n=a_{n,\mkern2mu n}A_{n-1} -a_{n,\mkern2mu n-1} a_{n-1,\mkern2mu n} A_{n-2}.$$ ($A_k$ denotes the leading principal minor of order $k$ of $A_n$). In the present case, you obtain the linear recurrence of order $2$: $$A_n=2A_{n-1}-A_{n-2},$$ with initial conditions: $A_1=2$, $\;A_2=3$.

Hint for the second determinant:

Let $I$ the unit matrix of order $n$, $U$ the $n\times n$ matrix with all coefficients $1$. The given matrix is but $I+2U$, and you can prove from multilinearity w.r.t. columns that $$\det A=\det I+2\sum \text{cofactors of }I=2n+1.$$

• Thank you, you gave me good perspective on the problem. – Marek Otypka Jan 1 '17 at 22:14