- #1
evinda
Gold Member
MHB
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Hello! (Wave)
Suppose that we are given this $(N+1) \times (N+1)$ matrix: $\begin{bmatrix}
-(1+h+\frac{h^2}{2}q(x_0)) & 1 & 0 & 0 & \dots & \dots & 0 \\
-1 & 2+h^2q(x_1) & -1 & 0 & \dots& \dots & 0\\
0 & -1 & 2+h^2q(x_2) & -1 & 0 & \dots & 0\\
& & & & & & \\
& & & & & & \\
& & & & & & \\
& & & & 0 & -1 & 2+h^2q(x_N)
\end{bmatrix}$ I want to show that the above matrix is invertible.So it suffices to show that the determinant is $\neq 0$, right?
Will the determinant be equal to $-\left( 1+h+\frac{h^2}{2}q(x_0) (2+h^2 q(x_1)) \cdots (2+h^2q(x_N)) +(2+h^2 q(x_2)) \cdots (2+h^2q(x_N)) \right)$?
Or am I wrong? (Thinking)
Suppose that we are given this $(N+1) \times (N+1)$ matrix: $\begin{bmatrix}
-(1+h+\frac{h^2}{2}q(x_0)) & 1 & 0 & 0 & \dots & \dots & 0 \\
-1 & 2+h^2q(x_1) & -1 & 0 & \dots& \dots & 0\\
0 & -1 & 2+h^2q(x_2) & -1 & 0 & \dots & 0\\
& & & & & & \\
& & & & & & \\
& & & & & & \\
& & & & 0 & -1 & 2+h^2q(x_N)
\end{bmatrix}$ I want to show that the above matrix is invertible.So it suffices to show that the determinant is $\neq 0$, right?
Will the determinant be equal to $-\left( 1+h+\frac{h^2}{2}q(x_0) (2+h^2 q(x_1)) \cdots (2+h^2q(x_N)) +(2+h^2 q(x_2)) \cdots (2+h^2q(x_N)) \right)$?
Or am I wrong? (Thinking)