Fundamentals of Matrix Algebra, 2011a

Chapter 2
Matrix Arithmec
is that there is a matrix B where BA = I, then we also know that AB = I.
That is, we know that B is the inverse of A (and hence A is inverble).
(c) We use the same logic as in the previous statement to show why this is the
same as “A is inverble.”
(d) If A is inverble, we can find the inverse by using Key Idea 10 (which in turn
depends on Theorem 5). The crux of Key Idea 10 is that the reduced row
echelon form of A is I; if it is something else, we can’t find A −1 (it doesn’t
exist). Knowing that A is inverble means that the reduced row echelon
form of A is I. We can go the other way; if we know that the reduced row
echelon form of A is I, then we can employ Key Idea 10 to find A −1 , so A is
inverble.
(e) We know from Theorem 8 that if A is inverble, then given any vector ⃗b,
A⃗x = ⃗b has always has exactly one soluon, namely ⃗x = A −1 ⃗b. However,
we can go the other way; let’s say we know that A⃗x = ⃗b always has exactly
soluon. How can we conclude that A is inverble?
Think about how we, up to this point, determined the soluon to A⃗x =
⃗b. We set up the augmented matrix [ A ⃗b ] and put it into reduced row
echelon form. We know that geng the identy matrix on the le means
that we had a unique soluon (and not geng the identy means we either
have no soluon or infinite soluons). So geng I on the le means having
a unique soluon; having I on the le means that the reduced row echelon
form of A is I, which we know from above is the same as A being inverble.
(f) This is the same as the above; simply replace the vector ⃗b with the vector
⃗0.
So we came up with a list of statements that are all equivalent to the statement
“A is inverble.” Again, if we know that if any one of them is true (or false), then
they are all true (or all false).
Theorem 9 states formally that if A is inverble, then A⃗x = ⃗b has exactly one solu-
on, namely A −1 ⃗b. What if A is not inverble? What are the possibilies for soluons
to A⃗x = ⃗b?
We know that A⃗x = ⃗b cannot have exactly one soluon; if it did, then by our theorem
it would be inverble. Recalling that linear equaons have either one soluon,
infinite soluons, or no soluon, we are le with the laer opons when A is not inverble.
This idea is important and so we’ll state it again as a Key Idea.
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