Lab #7 - Matrix Algebra and Chemometrics
PART A: Basic Matrix Calculations
1. Enter these two matrices:
A=
3
4
1
0
-1
2
3
-1
5
B=
1
-2
3
1
1
4
1
6
5
2. Give the result of adding (A + B) and multiplying (A*B) the two matrices together:
3. Now calculate B*A and write the results. Are the results of the two multiplications the
same? Is matrix multiplication commutative?
4. Calculate the transpose of matrix A:
5. Calculate the inverse of matrix A:
6. Calculate A*A-1 – what matrix is the result?
7. Calculate A-1*A. Is the result the same as in Step 6?
PART B: Solving Simultaneous Equations
8. Let's say that you have the following three simultaneous equations, and you want to
solve for x, y, and z x + 2y + 7z = 1
5x + y - z = 6
2x - 3y + 4z = 3
9. The problem can easily be solved using matrix methods. First recast the system of
equations in matrix form -
10. Solve for x, y, and z using an inverse and show your work here.
PART C: Linear Least Squares Revisited
11. Remember those complicated least squares equations you used in Program #5?
They are equivalent to the matrix solution to the problem, and of course using
matrices is so much easier! Here's how to do it. We have a data matrix Y, which
consists of a single row: Y = [y1 y2 y3 ....yn]
For example, this could be the data you fit before in the file ls.dat. Now, the equation
for a straight line is y = mx + b, but we're going to write it a little differently as:
Y = a o + a 1x
where ao = b and a1 = m.
The corresponding xaxis data is placed into a matrix X, where the first row of X is all
ones, and the second row is the xaxis data. The whole problem can be re-written as
Y=AX
Equation 1
Expanding the matrices results in this [y1 y2 y3 ....yn] = [ao a1]
1 1 1 ................1
x1 x2 x3.............xn
We want to solve for the fitting coefficients in matrix A, which can be done by using
the "right pseudoinverse". If X was a square matrix, the problem could be solved
simply by right multiplying both sides of Equation 1 by the inverse of X, so that the
solution would bea = Y  X-1
where the "-1" indicates taking the inverse.
But X is most definitely not square. So we right multiply both sides of Equation 1 by
the transpose of X, call it X' Y  X' = (a  X)  X'
which can be rewritten as -
Y  X' = a  (X  X')
because matrix multiplication is associative.
Now the matrix product (X  X') is just another matrix, so it will have an inverse, call it
(X  X')-1. We now right-multiply by this new inverse Y  X'  (X  X')-1 = a  (X  X')  (X  X')-1
Y  X'  (X  X')-1 = a
Equation 2 is the least squares solution to the problem.
which simplifies to
Equation 2
12. Load the file ls.dat into Matlab. Find the least squares best slope and y-intercept
using matrices. Plot the best-fit line over the data (plot the data as points and the fit
as a line). Print out the graph and hand-in with the exercise.
Enter values for b and m here: __________________________________
PART D: Linear Models
13. Consider a matrix D containing spectra of an acid-base indicator. Each column
contains the spectrum obtained at a particular pH. The pH of each spectrum
increases from left to right in the matrix. An example of this kind of data is shown in
Figure 1. Let's say that the matrix D has m rows and n columns. The dimensions of
matrix D and subsequent matrices can be found in Figure 2 on the next page.
3
2.5
In-
2
HIn
1.5
1
0.5
0
350
400
450
500
550
600
650
700
Figure 1
Acid-base indicators exist in two forms, a protonated and deprotonated species HIn

H+
+
InEquation 3
(acidic
form)
(basic
form)
The acidic form of the indicator is found as the left-hand peak in Figure 1; the
magnitude of this peak decreases with increasing pH. Conversely, the basic form
of the indicator is found as the right-hand peak; the magnitude of this peak
increases with increasing pH.
Now let's say that we have the spectra for the pure acidic form of the indicator (HIn)
and the basic form (In-). Place these spectra as columns in another matrix, call it
A. The dimensions of A are m rows by k columns as shown in Figure 2 (k = 2 in this
example but let's keep it more general).
Define a third matrix C such that the rows of C vary according to how the peaks
change in magnitude in the spectra. This matrix is called the design or model
matrix. In our example, the first row of C controls how the pure spectra found in
column 1 of matrix A changes. The second row of C controls the spectrum in
column 2 of matrix A. Matrix C will have dimensions k x n as shown in Figure 2.
Now comes the fun part. Matrix D is equal to the product of matrix A times C, it's
that simple! When matrix A is multiplied by C, each data point in D is a linear
combination of pure spectrum 1 plus pure spectrum 2. For example, in matrix D
element d11 = a11 x c11 + a12 x c21 (for the case where k = 2 again). The coefficients
c11 and c21 tell us how much of each pure component to add together, which
depends on the model we are using. The overall process is given by Equation 4 D= AC
d11 d12 ............ d1n
d21 d22 ............ d2n
.
.
.
.
.
.
.
.
.
.
dm1 dm2 ............ dmn
D=mxn
Equation 4
a11 a12 .......a1k
c1n
a21 a22 .......a2k
c2n
.
.
.
.
.
.
.
.
.
.
am1 am2 .....amk
A=mxk
c11 c12 .............
c21 c22 .............
.
.
.
.
.
.
ck1 ck2 ............. ckn
C=kxn
Figure 2
Usually you are trying to obtain matrix A because you are interested in obtaining
the spectra of the pure components of the mixture you are analyzing. Matrix A
can be solved for by using the matrix techniques we have been using already.
There is one major difference, and that is matrix C is usually never square. This
means that we cannot solve Equation 4 by right multiplying both sides by the
inverse of C. Instead we form the right pseudoinverse as follows. First right multiply
both sides by the transpose of C DC' = A (CC')
Then right multiply by the inverse of (CC') DC' (CC')-1 = A (CC')  (CC')-1
Since (CC')  (CC')-1 = I, the equation simplifies to A = DC' (CC')-1
Equation 5
This procedure is known as LINEAR MODELING. In general, linear modeling
consists of the following steps 1. Obtain data and place into the D matrix in column format.
2. Choose a model and place it into the C matrix in row format
3. Solve for A by using the right pseudoinverse
So how do you form matrix C in this problem? You place in the rows of C values
describing the variation in the pure spectral components. Typically there is some
experimental parameter being varied; in this example it is pH or the hydrogen ion
concentration. For acid-base indicators, the forms of the indicator vary according to
these expressions HIn = [H+]/([H+] + Ka)
In- = Ka/([H+] + Ka)
The term HIn gives the fraction of the indicator found in the acidic form; Inrepresents the fraction found as the basic form of the indicator. These functions are
known as the Distribution Function of the species.
With these background topics in mind, follow the procedure on the next page for
the acid-base indicator bromothymol blue. Bromothymol blue has a color change
from yellow to blue in the pH region 6.2-7.6. The pKa of this indicator is 7.10. We
are going to use linear modeling to extract the pure acid-base spectra of the
indicator.
14. The spectra of bromothymol blue in Figure 1 were taken at pH's of 6.06, 6.43, 7.04,
7.21, and 7.62. These data are contained in the file brom.txt which you can
download from the class webpage. The first column of this file contains the
wavelengths; the remaining columns contain the spectra in order of increasing pH.
15. Write a program that calculates the pure spectra of the acidic and basic forms of the
indicator, and plots (1) the spectra, (2) the C-matrix versus [H+], and (3) the A-matrix
containing the pure spectra.
16. Print out these graphs and hand them in with the lab.