Second difference matrices have beautiful inverses if they start with T1 = 1 (instead of K11 =2). Here is the 3 by 3 tridiagonal matrix T and its inverse: T-|-1 -1 2-1 0-1 2 One approach is Gauss-Jordan elimination on [TI]. I would rather write T as the product of first differences L times U. The inverses of L and U in Worked Example 2.5 A are sum matrices, so here are T = LU and T- = U-L-: 1 T = It' 1 0-1 difference 1 -1 [3 2 1] T= 2 2 1 1 1 1 T- = [:] sum difference Question. (4 by 4) What are the pivots of T? What is its 4 by 4 inverse? The reverse order UL gives what matrix T*? What is the inverse of T*? sum