|
- If
This can also be written as
This formula for the inverse of a 2 by 2 matrix is a good one to
memorize.
- No. Only square matrices can have an inverse such that
A-1A = AA-1 = I. However, there are one-sided inverses for some rectangular matrices, A, such that
AL-1A = I or AAR-1 = I. For example, if
This means that B is a right inverse for A. It is a "right"
inverse because A multiplied by B on the right produces an identity
matrix. It is "an" inverse rather than "the" inverse because it is not
unique. The matrix
is also a right inverse of A. A left inverse can be represented
by AL-1 and a right inverse can be represented by AR-1.
Remark 15 If someone asks if matrix A has an inverse, he or she is referring to a matrix A-1 such that A-1A = AA-1 = I. Therefore, unless the matrix A is square, just AB = I or BA = I is not sufficient proof that B is the inverse of A; it is only proof that B is at least a one-sided inverse of A. However, if the matrix is square and BA = I or AB = I, then B = A-1 and A-1A = AA-1 = I. You can find a proof of this
in a college linear algebra text.
- A lower triangular matrix has all the non-zero numbers on and below
the main diagonal. All the numbers above the main diagonal are zero. An
example is
Notice that there can be zeros on and/or below the main diagonal,
but all the numbers above the main diagonal MUST be zero.
- Yes, if the inverse exists, (A-1)T = (AT)-1. Let's look at a 2 by 2 matrix for an example.
Since cb = bc, these are equal. This only proves the 2 by 2 case. The
following is a proof for the general case:
|
AA-1
|
= |
I |
|
|
|
(AA-1)T |
= |
IT |
= |
I |
|
(A-1)TAT |
= |
I |
|
|
The last step follows because (AB)T = BTAT
as we showed in Chapter 3.
This proof proves our point because if AB = I, then
A is the inverse of B and B is the inverse of
A. Therefore, (A-1)T is
the inverse of AT.
- No. Try to substitute this answer into the original equation Ax = b.
You get
AAL-1b = b. This is only true for all b if A is square because AAL-1 ≠ I
if A is not square. However,
x = AR-1b is
always a solution to Ax = b because
AAR-1 = I.
- Yes, if there is a solution to Ax = b,
x = AL-1b will be the only
solution. Suppose that there were two solutions, x1 and x2, to Ax = b.
Then the following must be true:
|
Ax1 |
= |
b |
|
and |
|
Ax2 |
= |
b |
|
AL-1Ax1 |
= |
AL-1b |
|
and |
|
AL-1Ax2 |
= |
AL-1b | |
and
then
and
