Matrix Division
Je/ Stuart
c
2008
ax
=
b
where
a
and
b
are real numbers.
Why do we need division? In its earliest form, division must have arisen to
answer questions such as, "If our foraging party of three gathered seven melons,
3
x
= 7
where
x
which is to say, in solving problems of the form
ax
=
b
where
a
and
b
are known
numbers. Of course, we all learned early in our educations that
ax
=
b
has a
unique solution when
a
6
= 0
, namely,
x
=
b=a
.
Although this seems trivially
ax
=
b
. We
have observed that when
a
6
= 0
, there is a unique solution for every possible
choice of
b
, namely
x
=
b=a
. What happens when
a
= 0
? Probably, your ±rst
thought is that there is no solution since you cannot divide by zero.
Is this
correct? Certainly, if
b
6
= 0
;
we cannot ±nd a real number
x
such that
0
x
6
= 0
.
In the language of linear systems, when
a
= 0
and
b
6
= 0
, the system
ax
=
b
is inconsistent.
But what happens when
b
= 0
?
In this case, the equation
becomes
0
x
= 0
, which holds for every real number
x
.
That is, when
a
= 0
and
b
= 0
, the system
ax
=
b
has in±nitely many solutions. Summarizing:
Theorem 1
Let
a
and
b
be real numbers.
±
If
a
= 0
, then for all real numbers
b
with
b
6
= 0
,
ax
=
b
is inconsistent;
and for
b
= 0
,
ax
=
b
±
If
a
6
= 0
, then for all real numbers
b
,
ax
=
b
is consistent and has a unique
solution,
x
=
b=a
.
What happens for linear systems
Ax
=
b
with more equations and
more variables?
From the Trichotomy Theorem, we know that exactly one of three outcomes
occurs for the linear system
Ax
=
b
where
A
and
b
are known: the system is
inconsistent, the system is consistent with a unique solution, or the system is
consistent with in±nitely many solutions. Is there some analogous property to
a
6
= 0
that distinguishes the situation in which there is a unique solution to
Ax
=
b
when
Ax
=
b
is solvable, or that distinguishes the situation in which
Ax
=
b
is solvable for all vectors
b
?
When do multiple solutions occur for
Ax
=
b
? Recall that when we form
the product
Ax
, we are creating a vector that is a linear combination of the
columns of
A
using the scalars that are the entries of
x
. One consequence of
this is that for any matrix
A
, we can always ±nd a vector
b
for which
Ax
=
b
is
1
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View Full Documentconsistent; just choose any vector
b
that is a linear combination of the columns
of
A
, or even simpler, just choose
b
to be one of the columns of the matrix
A
.
Also recall that for a consistent system, multiple solutions occur exactly when
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 Spring '08
 Neely
 Linear Algebra, ax, Invertible matrix

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