The very last words say that the 5 by 5 centered difference matrix is not invertible

Find the linear combination 2s 1 + 3s 2 + 4s 3 = b. Then write b as a matrix-vector multiplication S x. Compute the dot products (row of S) . x: 81 = [:] 82 = [n 83 = [n go into ilie columns of S.

Solve these equations S y = b with s 1, S2, S3 in the columns of S:

Solve these three equations for Yl, Y2, Y3 in terms of B1, B2 , B3: Sy = B Write the solution y as a matrix A = S-l times the vector B. Are the columns of S independent or dependent?

Find a combination XI W 1 + X2 W2 + X3 W 3 that gives the zero vector: Those vectors are (independent) (dependent). The three vectors lie in a . The matrix W with those columns is not invertible.

The rows of that matrix,W produce three vectors (I write them as columns): \ Linear algebra says that these vectors must also lie in a plane. There must be many combinations with Ylrl + Y2r2 + Y3r3 = O. Find two sets of y's.

Which values of c give dependent columns (combination equals zero)? o c] 1 0 1 1 [ c c c]

If the columns combine into Ax = 0 then each row has r . x = 0: =Ul The three rows also lie in a plane. Why is that plane perpendicular to x?

Moving to a 4 by 4 difference equation Ax = h, find the four components XI, X2, X3, X4. Then write this solution as x = Sh to find the inverse matrix S = A-I:

What is the cyclic 4 by 4 difference matrix C? It will have I and -1 in each row. Find all solutions x = (Xl, X2, X3, X4) to ex = O. The four columns of C lie in a "three-dimensional hyperplane" inside four-dimensional space.

Aforward difference matrix I::J. is upper triangular: [ -1 1 0] [ ZI ] [ Z2 - Zl ] [ b l ] I::J.z = 0 -1 1 Z2 = Z3 - Z2 = b2 = h. o 0 -1 Z3 0 - Z3 b3 Find ZI, Z2, Z3 from bl , b2 , b3. What is the inverse matrix in z = I::J. -1 h?

Show that the forward differences (t + 1)2 - t 2 are 2t+ 1 = odd numbers. As in calculus, the difference (t + l)n - tn will begin with the derivative of tn, which is _

The last lines of the Worked Example say that the 4 by 4 centered difference matrix in (16) is invertible. Solve ex = (b l , b2 , b3, b4 ) to find its inverse in x = c-l h.

The very last words say that the 5 by 5 centered difference matrix is not invertible. Write down the 5 equations ex = h. Find a combination of left sides that gives zero. What combination of bl , b2 , b3, b4 , bs must be zero? (The 5 columns lie on a "4-dimensional hyperplane" in 5-dimensional space.)

If (a,b) is a multiple of (c,d) with abcd =I- 0, show that (a,c) is a multiple of (b, d). This is surprisingly important; two columns are falling on one line. You could use numbers first to see how a, b, c, d are related. The question will lead to: The matrix A = [~ ~] has dependent columns when it has dependent rows.