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

Question

anonymous · Answer

See the attachment.

anonymous · Answer

@wio

anonymous · Answer

Is that all the info they are giving?

anonymous · Answer

Yes.

anonymous · Answer

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

anonymous · Answer

how did you get that?

anonymous · Answer

i think by hand.

anonymous · Answer

But isn't the inverse of 0 is 0?

anonymous · Answer

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

anonymous · Answer

Can you help me to find the inverse?

anonymous · Answer

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?

anonymous · Answer

@precal

anonymous · Answer

yes. there is a correction. $$z_1=-b_1-b_2-b_3$$

precal · Answer

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