Answer
See solution
Work Step by Step
a)
Given:
$(I-C) x=d$ and $(I-C) \Delta x=\Delta d$
If the final demand changes to $d+\Delta d,$ then :
$d+\Delta d=(I-C) x+(I-C) \Delta x=(I-C)(x+\Delta x)$
Meaning that the production level must change to $x+\Delta x$
2
b)
If $(I-C) \Delta x=\Delta d,$ then $\Delta x=(I-C)^{-1} \Delta d$
Now, if $\Delta d$ has 1 in first row and 0 everywhere else, by matrix multiplication property, result $\Delta x$ will be first column of $(I-C)^{-1}$
It is like multiplying a matrix with first column of $I$ - only first column of matrix survives.
