User Manual

90 Section 4: Using Matrix Operations
90
Multiplying the calculated inverse and the original matrix verifies that the calculated inverse
is poor.
The trouble is that E is badly scaled. A well-scaled matrix, like A, has all its rows and
columns comparable in norm and the same must hold true for its inverse. The rows and
columns of E are about as comparable in norm as those of A, but the first row and column of
E
−1
are small in norm compared with the others. Therefore, to achieve better numerical
results, the rows and columns of E should be scaled before the matrix is inverted. This means
that the diagonal matrices L and R discussed earlier should be chosen to make LER and
(LER)
−1
= R
−1
E
−1
L
−1
not so badly scaled.
In general, you can't know the true inverse of matrix E in advance. So the detection of bad
scaling in E and the choice of scaling matrices L and R must be based on E and the
calculated E
−1
. The calculated E
−1
shows poor scaling and might suggest trying












5
5
5
1000
0100
0010
RL
.
Using these scaling matrices,















3030
3030
10
10102
10101
21103
LER
,
which is still poorly scaled, but not so poorly that the HP-15C can’t cope. The calculated
inverse is
















3030
3030
30
1
101021
1021043
13102
)(LER
.
This result is correct to 10 digits, although you wouldn't be expected to know this. This result
is verifiably correct in the sense that using the calculated inverse,
(LER)
−1
(LER) = (LER)(LER)
−1
= I (the identity matrix)
to 10 digits.
Then E
−1
is calculated as

















4040
4040
40
11
101021
1021043
13102
)( LLERRE
,
which is correct to 10 digits.