On page 18-12 of the HP48G User's Guide, it says:

"Be sure to pay attention to the nature of the linear system you are solving because it will influence how you should interpret the result array. In some cases, you will want to test for ill-conditioning before accepting even an exact solution as a "true" solution."

I will give a system that exemplifies a situation where the exact solution should probably not be considered the "true" solution.

Imagine that you need to measure an electric current, and to increase accuracy you will make the measurement with 4 nominally identical meters and derive some kind of weighted average of the readings of the 4 meters. These meters are just simple analog panel meters.

To get a weighting factor for each meter, you connect the 4 meters in series and pass an accurately known current through all 4 meters at the same time. Pass a current of 1, 2, 3 and 4 amperes through the meters, and note the reading on each meter. Create a matrix of the meter readings and set up a linear sytem like this:

A*W = B

A is the matrix of meter readings and B is the column matrix of the accurate currents that were passed through the meters in series. W will be the weighting factors to be applied to the meter readings in the future when this arrangement is used to measure currents. Each column is the 4 readings from a particular meter for the 4 currents. Each row is the readings from the 4 meters for a particular current.

The matrix A of readings is:

[[ 0.95 1.09 1.01 1.09 ]

[ 1.94 1.95 2.02 1.91 ]

[ 2.92 3.00 2.93 2.98 ]

[ 4.01 3.95 3.97 3.94 ]]

The column matrix of currents is:

[[ 1 ]

[ 2 ]

[ 3 ]

[ 4 ]]

What we want to do is get a vector W which will contain 4 numbers, weighting factors to be applied to the meter readings. In other words, if the 4 meter readings are m1, m2, m3 and m4, and the weighting factors (elements of W) are w1, w2, w3 and w4, then the derived current value will be w1*m1+w2*m2+w3*m3+w4*m4. Obviously, given that the 4 meters are nominally identical, weighting factors of w1=.25, w2=.25, w3=.25 and w4=.25 are close to what we want. Since the meters will have some unavoidable manufacturing tolerances, the weighting factors will be slightly perturbed from the ideal values I just gave.

Other solutions to the problem would be W vectors of (the T means transpose):

[ 1 0 0 0 ]T, just use the readings from the 1st meter.

[ 0 1 0 0 ]T, just use the readings from the 2nd meter, etc.

[ .5 .5 0 0 ]T, just use the readings of the 1st and 2nd meters.

[ .5 0 .5 0 ]T, just use the readings of the 1st and 3rd meters, etc.

If we solve the A*W = B linear system on the HP48G (or its descendants), we get:

[[ .8822598871 ]

[ 3.1448587566 ]

[ -.4836158193 ]

[ -2.5482485873 ]]

While this is an exact mathematical solution, it doesn't make sense given the physical situation. Two of the weights are negative! And the system isn't particularly ill-conditioned; the condition number of the A matrix is about 1000, so we lose about 3 digits of accuracy in the solution. What we would like is 4 weights of about .25 each.

Here is a case where the exact mathematical solution doesn't make sense. It reminds me of those freshman physics problems where you get two solutions to a problem, one of which doesn't fit the physical situation.

What shall we do?