# Is there anything like upper tridiagonal matrix? How to find the determinant of such a matrix?

I want to find the determinant of the following matrix.

$$\left[\begin{matrix} -\alpha_1 & \beta_2 & -\gamma_3 & 0 & 0 & 0 & \cdots & 0&0 \\ 0 & -\alpha_2 & \beta_3 & -\gamma_4 & 0 & 0 & \cdots & 0 & 0 \\0 & 0 & -\alpha_3 & \beta_4 & -\gamma_5&0&\cdots&0 & 0 \\\vdots &\vdots&\ddots &&&&&\vdots&\vdots \\0&0&0&0&0&0&\cdots&\beta_{N-2}&-\gamma_{N-1} \\0&0&0&0&0&0&\cdots&-\alpha_{N-2}&\beta_{N-1}\end{matrix}\right]$$

Is there any closed form formula for the determinant of such a matrix?

Or can I establish some recurrence relation for the determinant, like the tridiagonal matrix?

• How do you define the determinant of a non-square matrix? If it is a mistake and the matrix has as las element $\alpha_{N-2}$, then the determinant is simply the product of the elements in the diagonal. This holds for any upper triangular matrix. – Darío G Nov 7 '16 at 15:45

The matrix that you put in the question is not square since it has size $(N-2)\times (N-1)$ (you can check by counting the $\alpha's$ in each row and column).

If your matrix is supposed to have the form

$$T_n=\begin{bmatrix}a_1 & b_1 & c_1 & 0 & \cdots &0\\ 0 & a_2 & b_2 & 0 & \cdots&0\\ 0 & 0 & a_3 & b_3 &\cdots & 0\\ \vdots & \vdots &\vdots &\ddots & \cdots & 0\\ 0 & 0 & \cdots & & a_{n-1} & b_{n-1}\\ 0 & 0 &\cdots & & 0 & a_n\end{bmatrix}$$ then it is simply an upper triangular matrix and the determinant is $\displaystyle{\det(T_n)=\prod_{i=1}^n a_i}$.

If your matrix has the form:

$$A_n=\begin{bmatrix}a_1 & b_1 & 0 & 0 & \cdots &0\\ c_1 & a_2 & b_2 & 0 & \cdots&0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0\\ \vdots & \vdots &\vdots &\ddots & \cdots & 0\\ 0 & 0 & \cdots & & a_{n-1} & b_{n-1}\\ 0 & 0 &\cdots & & c_{n-1} & a_n\end{bmatrix}$$

Then I can provide a recursive relation. To start, notice that \begin{align*} \det(A_1)&=a_1, \text{ and } \\ \det(A_2)&=a_1a_2-b_1c_1 \text{ as expected.}\\ &\\ \det(A_3)&=a_3\cdot \det(A_2)-a_1b_2c_2\\ &=a_1a_2a_3-b_1c_1a_3-a_1b_2c_2\\ &=a_3\cdot \det(A_2)-b_2c_2\cdot \det(A_1) \end{align*}

Now, for the matrix $A_{n+1}$ we have

\begin{align*} \det(A_{n+1})&=\begin{vmatrix}a_1 & b_1 & 0 & 0 & \cdots &0 & 0\\ c_1 & a_2 & b_2 & 0 & \cdots&0 & 0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0 & 0\\ 0 & 0 & c_3 & \ddots & \ddots & 0 & 0 \\ \vdots & \vdots &\vdots &\ddots & a_{n-1} & b_{n-1} & 0 \\ 0 & 0 & \cdots & 0 & c_{n-1} & a_{n} & b_{n}\\ 0 & 0 &\cdots & 0 & 0 & c_{n} & a_{n+1}\end{vmatrix}\\ &=(-1)^{(n+1)+(n+1)}\cdot a_{n+1}\cdot \begin{vmatrix}a_1 & b_1 & 0 & 0 & \cdots &0\\ c_1 & a_2 & b_2 & 0 & \cdots&0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0\\ 0 & 0 & c_3 & \ddots & \ddots & 0 \\ \vdots & \vdots &\vdots &\ddots & a_{n-1} & b_{n-1}\\ 0 & 0 & \cdots & 0 & c_{n-1} & a_{n}\end{vmatrix}\\ &\\ & + (-1)^{(n+1)+n}\cdot c_{n-1}\cdot \begin{vmatrix}a_1 & b_1 & 0 & 0 & \cdots & 0\\ c_1 & a_2 & b_2 & 0 & \cdots & 0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0\\ 0 & 0 & c_3 & \ddots & \ddots & 0 \\ \vdots & \vdots &\vdots &\ddots & a_{n-1} & 0 \\ 0 & 0 & \cdots & 0 & c_{n-1} & b_{n-1}\end{vmatrix}\\ &\\ &=a_{n+1}\cdot \det(A_n)-c_{n}b_{n}\det(A_{n-1}). \end{align*}

So, the sequence of determinants $\{x_n=\det(A_n):n\in\mathbb{N}\}$ can be defined recursively by

$$\begin{cases}x_0=1\\ x_1=a_1\\ x_{n+1}=a_{n+1}x_n-b_nc_n\cdot x_{n-1}\end{cases}$$

I asked in this question if some closed form can be obtained. So far, I can say that in general we have

$$\begin{bmatrix}x_{n+1}\\ x_n\end{bmatrix}=\left(\begin{bmatrix}a_{n+1} & -b_{n}c_{n}\\ 1 & 0\end{bmatrix}\cdots \begin{bmatrix}a_3 & -b_2c_2\\ 1 & 0\end{bmatrix}\cdot \begin{bmatrix}a_2 & -b_1c_1\\ 1 & 0\end{bmatrix}\right)\cdot \begin{pmatrix}a_1\\ 1\end{pmatrix}$$

but it feels kind of ironic to me that to find the determinant we "only" have to multiply these $n$ matrices