Question

Please help me with linear algebra problem? (The number after the letters z and b are subscripts.)

Answer 1

See the attachment.

Answer 2

@wio

Answer 3

Is that all the info they are giving?

Answer 4

Yes.

Answer 5

There's 2 parts. We first need to find z1, z2, z3. And the inverse matrix in z=(triangle symbol)^-1 * b.

Answer 6

how did you get that?

Answer 7

i think by hand.

Answer 8

But isn't the inverse of 0 is 0?

Answer 9

from the last equality relation, we have
\[
\begin{align}
z_2-z_1&=b_1\\
z_3-z_2&=b_2\\
-z_3&=b_3\\
\implies z_2&=-b_2-b_3\\
\implies z_1&=b_2+b_3-b_1
\end{align}
\]
then they ask for the inverse of the forward difference matrix \({\bf \Delta}\)
solve for inverse of \[
\begin{bmatrix}
-1&1&0\\0&-1&1\\0&0&-1
\end{bmatrix}
\]
since it is already upper triangular matrix, you can use the short algorithms like Gauss-Jordan elimination method

Answer 10

Can you help me to find the inverse?

Answer 11

What I got from z1, z2 and z3 are different from the answers above. I got z1=-b3-b2-b1, z2=-b3-b2, and z3=-b3. How come?

Answer 12

@precal

Answer 13

yes. there is a correction. \[z_1=-b_1-b_2-b_3\]

Answer 14

sorry not up to date in linear algebra. I would just use the calculator - I know that is the easy way out.......