1

Ultraviolet – Visible Spectroscopy for Determination of α- and

β-acids in beer hops

Introduction

In this lab, you will undertake a simple extraction of two hop samples followed by

spectrophotometric analysis at three wavelengths to quantify the quantities of α- and β-acids,

as well as a third component linked to hop degradation.

UV-Vis spectroscopy (or spectrophotometry) is one of the most important quantitative

spectroscopic techniques available in the modern arsenal of analytical instrumentation for

molecular analysis. The wavelength range extends from about 190 – 750 nm, which

corresponds to molecular electronic transitions of different origins. Consider the electronic

energy level diagram of a typical molecule (Figure 1):

The four transitions correspond to:

1. σ – σ* transitions

This transition type requires large energy, which may result in bond breaking. It is not very

important from an analytical point of view. This type of transition typically requires energetic

photons, below 190 nm.

2. π – π* transitions

This is the most analytically useful type of transition. The energy required for this type of

transition is moderate, corresponding to photons in the 190 – 750 nm range.

3. n – π* and n – σ* transitions

Figure 1: Energy level diagram

representing electronic transitions for

a typical molecule

2

Molecules containing lone pair(s) of electrons exhibit some special properties. Electrons that do

not participate in chemical bonds can absorb energy and are excited either to the π* (if the

molecule has π – bonds) or σ* state. Unfortunately, these transitions are also not very useful

analytically.

Spectrophotometric absorbance features in the UV-Vis tend to be quite broad. That is, the

molecule absorbs significantly over a wide range of wavelengths. For example, the absorbance

spectra of two dyes commonly used in lasers are shown in Figure 2.

Beer-Lambert Law

The analytical utility of UV-vis photon absorption derives from the (intuitive) fact that the

number of photons absorbed by an absorbing species depends on the number of absorbing

molecules and, therefore, on its concentration. The Beer-Lambert Law states that the

concentration of an absorbing species is logarithmically related to the fraction of light

transmitted (T): this is termed the “absorbance, A.” So, A = -log (T). But, you would also expect

the absorbance to depend on the “strength of the absorber,” as well as the “thickness of the

absorber.” For example, if you had a 1-mm thick piece of light absorbing plastic, you would

expect that a plastic piece that is twice as thick would absorb twice as much light. So, putting

this all together, the Beer-Lambert Law states

𝐴 = − log(𝑇) = ε𝑏𝐶

where ε is the molar absorptivity (M-1 cm-1) or “strength of the absorber”. Note that ε could

also be expressed in L/g if concentration is expressed in g/L. This is called the specific

absorptivity

b is the pathlength (cm) or “thickness of the absorber”

C is the concentration (M-1) of the absorber

Figure 2. Absorbance spectrum of Nile Blue and Coumarin 30 in

ethanol. Note that absorbance is given in arbitrary units.

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

200 300 400 500 600 700

Ab

sorb

an

ce (a

.u.)

Wavelength (nm)

Absorbance of dyes in ethanol

Nile Blue

Coumarin 30

3

The molar absorptivity is the most important indicator of analytical sensitivity in UV-Vis. When

“b” is 1-cm (most common) and A is plotted as a function of concentration, a straight line

results with a slope equal to the molar absorptivity. So, in effect, the molar absorptivity is the

change absorbance per unit change in concentration.

As with any other instrumental method of analysis, the response must be calibrated for analyte

concentration and this is the most common approach for samples that do not present any

interference with measurement of the absorbance. Owing to the broad absorption spectra

exhibited by most solvated molecules, one of the most common interferences in UV-Vis

spectroscopy comes from the presence of multiple absorbers in the same sample. For example,

is a sample contained only Nile Blue and Coumarin 30, then calibration curves could be derived

for each dye at its wavelength of maximum absorbance (without fear of the absorbance by one

of the dyes interfering with measurement of the other dye). If, on the other hand, the sample

also contained Coumarin 1 (Figure 3), then it is clear that independent calibration curves could

not be developed for the two Coumarin dyes, as their absorbance spectra overlap in the range

350 – 450 nm. It would therefore be impossible to measure independently the absorbance of

one of the dyes in the presence of the other.

Fortunately, absorbance is an additive property, and so the total absorbance (AT) and any single

wavelength is equal to the sum of the absorbance by all absorbers at that wavelength:

𝐴𝑇(𝜆1) = ∑ 𝐴𝑖(𝜆1)

𝑖

= ∑ ε𝑖(𝜆1)𝑏𝐶𝑖

𝑖

𝐴𝑇(𝜆2) = ∑ 𝐴𝑖(𝜆2)

𝑖

= ∑ ε𝑖(𝜆2)𝑏𝐶𝑖

𝑖

Figure 3. Absorbance spectrum of Nile Blue, Coumarin 30 and Coumarin 1 in ethanol. Note that absorbance is given in arbitrary units.

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

200 300 400 500 600 700

Ab

sorb

ance

(a.

u.)

Wavelength (nm)

Absorbance of dyes in ethanol

Nile Blue

Coumarin 30

Coumarin 1

4

Therefore, by measuring the total absorbance at the absorption maximum of each component,

and knowing the “strength of absorption” of each component at each wavelength, we can solve

the set of equations simultaneously to determine the concentration of each absorber.

Analysis of α- and β-acids in beer hops

Hops contain both humulones (α-acids) and lupulones (β-acids), the major chemical

components of which are shown in Figure 4.

It is important for a brewer to know the quantity of α–acids in the hop used because α–acids

isomerize to form iso-α–acids during the brewing process, adding bitterness to balance the

flavor of the finished beer.

Materials

You will be provided with two samples of dried, commercial hops used for home brewing.

Additionally, you will use reagent grade NaOH and spectrophotometric grade methanol for the

acids extraction.

The spectrophotometer you will use for this experiment is the Shimadzu UV2450 double beam

instrument. Operating instructions are provided on the Chem 219 BlackBoard site.

Hazards

Methanol is highly flammable and toxic by inhalation, ingestion, or skin absorption. Waste

methanol should be placed in a labeled non-halogenated waste container for disposal (see your

TA). Sodium hydroxide (NaOH) is corrosive and can cause severe burns. Wear suitable personal

protection equipment.

Procedure: (To be conducted in triplicate)

Figure 4. Structures of major α - and β-acids found in hops.

5

Prepare 250 mL of methanolic NaOH by mixing 0.5 mL of 6 M NaOH into 250 mL methanol.

Grind hop pellets approximately 1 gram at a time (otherwise you will clog the grinder) and

combine until you have approximately 8-9 g of each type of hop.

Pipette exactly 50.0 mL of methanol into a 100 mL beaker. Accurately weigh out (to the nearest

mg) 2.5 g of ground hops and add to the beaker containing the methanol. Add a stir bar, place

on a magnetic stir plate and stir for 30 min at room temperature. Allow to stand without

stirring for 10 min to let the particulate matter settle. Gravity filter the settled solution into a

separate, clean, dry 125-mL Erlenmeyer flask. [Note: Do not use additional solvent to wash the

beaker or the filtered solid (why?)]. Pipette a 50 μL aliquot of the filtrate into a 25-mL

volumetric flask and bring up to volume with the methanolic NaOH you prepared.

Fill a 1-cm quartz cuvette approximately ¾-full with blank solution (this is the methanolic NaOH

solution you used to prepare all your samples) and place it in the reference side cell holder.

Rinse a separate cuvette several times with your hops extract and finally fill the 1-cm quartz

cuvette about ¾-full with an extract solution. Place this cuvette in the sample cell holder.

Measure a background corrected absorbance spectrum of the extract solution from 210 nm to

510 nm.

Data analysis

Because there are two major components in the hop extract, the mixture of α- and β-acids

appears to be an ideal system for the classic two-component analysis covered in most

quantitative analysis texts. All that is needed is the molar absorptivity coefficients at two

different wavelengths for each component, so a system of two equations can be solved for two

unknowns. The “fly in the ointment, or beer in this case” is that the hops natural product

extract is more complex. Firstly, neither the α-acids nor the β-acids are single compounds (see

Fig. 4). Additionally, there is a third component that appears over time as the α- and β- acids

are degraded. This third component has not been purified and is thought to be some other

breakdown component of the hops. It absorbs strongly at 275 nm but has significant

absorptions at 325 and 355 nm, where it augments the absorption of α- and β-acids and

interferes with a standard two-component analysis.

Because we are not dealing with single absorbing compounds, it is more practical to use specific

absorptivity than molar absorptivity. Specific absorptivity relates the absorbance measurement

to that of a mixture of compounds with a total concentration of 1 g/L at a given path length

(typically 1 cm). Alderton et al. (Alderton, Bailey et al. 1954) reported specific absorptivities (L g-

1 cm-1) for pure α- and β-acids, as well as the third degradation component. These are given in

Table 1.

355 nm 325 nm 275 nm

α-acids 31.8 38.1 9.0

6

β-acids 46.0 33.1 3.7

Comp3 1.0 1.5 3.1

Table 1: Specific absorptivities (L g-1 cm-1) for α- and β-acids in beer hops, as well as the third

absorbing component (see Ref. 1).

Because the total absorbance at any wavelength is equal to the sum of absorbances by each

component, we can describe this three-component system with three equations:

𝐴355 = 31.8𝐶𝛼 + 46.0𝐶𝛽 + 1.0𝐶𝐶𝑜𝑚𝑝3

𝐴325 = 38.1𝐶𝛼 + 33.1𝐶𝛽 + 1.5𝐶𝐶𝑜𝑚𝑝3

𝐴275 = 9.0𝐶𝛼 + 3.7𝐶𝛽 + 3.1𝐶𝐶𝑜𝑚𝑝3

Although this system of three equations can be solved by any graphing calculator (no duh), we

will use the method of Gaussian elimination (which is what your calculator is doing under the

hood). The method of Gaussian elimination can be used to solve virtually any system containing

an equal number of unknowns and equations. An introduction to the Gaussian elimination

method is given in the Appendix.

Of course, the brewer is not interested in knowing the concentration of α- and β-acids in an

extract, but rather wants to know the percentage of α- and β-acids in the actual dried hop. So,

your final task is to work through your procedure and dilutions to find this number. Be sure to

report appropriate statistics for your results.

Analysis point 1. From the overall shape of your absorbance curve, decide if your sample contains mostly humulones (alpha-acids) or lupulones (beta acids).

Analysis point 2. Make a data table containing the absorbance values for all solutions at 355 nm, 325 nm, and 275 nm. Use the data to do a two component analysis of your spectra and determine the grams of humulones and lupulones in your extract and the % alpha and beta acids in your original hops sample.

Analysis point 3. Use a three-component analysis to find the concentration of α- and β- acids in the solutions you analyzed. How do these values compare to your two-component analysis? Next, use your values from the three component analysis to determine the concentration of α- and β- acids in the actual extract. Finally, determine the % α- and β- acids in the original hops sample. How do they compare to your two-component analysis?

Alderton, G., et al. (1954). "Spectrophotometric Determination of Humulone Complex and Lupulone in

Hops." Anal. Chem. 26: 983 - 992.

Appendix: Mathematical Background using the Hops Data as a Real World Example

7

The hops lab provides an opportunity to solve a system of linear equations with algebra and

matrices. Here you can practice methods using real-world numbers rather than the artificial

whole numbers that are typically used in mathematics texts, designed to work without fractions

or decimals.

The absorbance of a hops solution at three key wavelengths can be found (assuming a 1 cm path

length through the solution) by the equations:

A355 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3

A325 = 38.1 Cα + 33.1 Cβ + 1.5 CComp3 The absorption equations

A275 = 9.0 Cα + 3.7 Cβ + 3.1 CComp3

Using sample hops absorption, where A355 = 0.55, A325 = 0.51, and A275 = 0.29, the absorption

equations become:

0.55 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3

0.51 = 38.1 Cα + 33.1 Cβ + 1.5 CComp3 The absorption equations for hops

0.29 = 9.0 Cα + 3.7 Cβ + 3.1 CComp3

The concentrations of α-acids (Cα), β-acids (Cβ) and the break down component (CComp3 ) in the

hops solution can be determined from these absorption equations. Here we present some methods

for solving this system of linear equations, including:

A. Gaussian elimination with back substitution using an algebraic approach, an equivalent

matrix representation, and a matrix row-echelon method.

B. Gauss-Jordan elimination, a matrix technique that does not require back substitution.

C. Solving directly from concentration equations derived from the absorption equations. The

concentration equations are derived using the inverse matrix, which is found with a

system of linear equations, Gauss-Jordan elimination of adjoined identity and coefficient

matrices, a TI calculator, and Excel.

D. Matrix math on a TI calculator.

E. Matrix math in Excel.

A. Solving for Cα, Cβ, and CComp3 With Gaussian Elimination with Back Substitution

1. Algebraic Method

Gaussian elimination with back substitution works well for solving systems of linear equations

by hand. In this technique the equations are reduced by a series of elementary operations to one

equation with one unknown, whose solution is then used through back-substitution to find the

other unknowns.

8

For the Glacier hops solution, the three absorption equations to be solved are:

0.55 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3 EQN 1

0.51 = 38.1 Cα + 33.1 Cβ + 1.5 CComp3 EQN 2

0.29 = 9.0 Cα + 3.7 Cβ + 3.1 CComp3 EQN 3

In this technique, EQN 1 remains unchanged throughout. One variable is selected to be

eliminated from EQN 2, say Cα, by first multiplying EQN 2 by the ratio of (the Cα coefficient

from EQN 1/the Cα coefficient from EQN 2)…

(31.8

38.1) (0.51 = 38.1 Cα + 33.1 Cβ + 1.5 CComp3)

or 0.4257 = 31.8 Cα + 27.627 Cβ + 1.2520 CComp3

…and then subtracting this result from EQN 1:

0.55 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3

‒ (0.4257 = 31.8 Cα + 27.627 Cβ + 1.2520 CComp3)

0.1243 = 0 + 18.373 Cβ – 0.2520 CComp3 EQN 2′

Two variables, Cα and Cβ, are next eliminated from EQN 3 in two steps. In step 1, Cα is

eliminated the same way as before, by multiplying EQN 3 by the ratio of (the Cα coefficient from

EQN 1/the Cα coefficient from EQN 3)…

(31.8

9.0) (0.29 = 9.0 Cα + 3.7 Cβ + 3.1 CComp3)

or 1.0247 = 31.8 Cα + 13.073 Cβ + 10.953 CComp3

…and then subtracting this result from EQN 1:

0.55 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3

‒ (1.0247 = 31.8 Cα + 13.073 Cβ + 10.953 CComp3)

-0.4747 = 0 + 32.927 Cβ – 9.953 CComp3 EQN 3intermediate

In step 2, Cβ is eliminated from this intermediate equation by first multiplying EQN 3intermediate by

the ratio of (the Cβ coefficient from EQN 2′/the Cβ coefficient from EQN 3intermediate)…

(18.373

32.927)(-0.4747 = 0 + 32.927 Cβ – 9.953 CComp3)

9

or -0.2649 = 0 + 18.373 Cβ – 5.554 CComp3

…and then subtracting this result from the EQN 2′:

0.1243 = 0 + 18.373 Cβ – 0.2520 CComp3

-(-0.2649 = 0 + 18.373 Cβ – 5.554 CComp3)

0.3892 = 0 + 0 + 5.302 CComp3 EQN 3′

The three absorption equations are now:

0.55 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3 EQN 1

0.1243 = 0 + 18.373 Cβ – 0.2520 CComp3 EQN 2′

0.3892 = 0 + 0 + 5.302 CComp3 EQN 3′

EQN 3′ is one equation with one unknown, so CComp3 can be found by:

CComp3 = (0.3892

5.302) = 0.07341 grams/liter

Cβ can now be found by back-substituting the value for CComp3 into EQN 2′:

0.1243 = 0 + 18.373 Cβ – 0.2520 (0.07341)

Cβ = (0.1243 + (0.2520)(0.07341))/18.373 = 0.007772 grams/liter

Cα can now be found by back-substituting the values for Cβ and CComp3 into EQN 1:

0.55 = 31.8 Cα + 46.0 (0.007772) + 1.0 (0.07341)

Cα = (0.55 – (46.0)(0.007772) – (1.0)(0.07341))/31.8 = 0.003745 grams/liter

Accounting for significant digits then yields the values shown in Table 1:

Cα = 0.0037 g/L

Cβ = 0.0078 g/L

CComp3 = 0.073 g/L

10

2. Matrix Representation of the Algebraic Method

The three absorption equations can be written in matrix form as:

Putting in the absorption values for the Glacier hops solution, the matrix can be written in

simplified form as:

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

⋮ ⋮ ⋮

0.550.510.29

]

The augmented matrix can be solved using the same steps as the algebraic method, but the

matrix representation is more streamlined as only the coefficients need to be written.

The first row of the matrix remains unchanged. The first term in the second row is then

converted to 0 with elementary row operations:

The first term in the third row is then converted to 0:

The second term in the third row is then converted to 0:

R3′ is used to find CComp3 = 5.30/0.389 = 0.0734. Cβ can be found using back substitution into R2′

as in the algebraic method. Cα can then be found using back substitution into R1.

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

] [

𝐶𝛼

𝐶𝛽

𝐶3𝑟𝑑

] = [

𝐴355

𝐴325

𝐴275

]

R1, equivalent to EQN 1

R2, equivalent to EQN 2

R3, equivalent to EQN 3

[31.8 46.0 1.0

0 18.37 −0.2529.0 3.7 3.1

⋮ ⋮ ⋮

0.550.1240.29

]

R2′= R1 – (31.8/38.1)R2 , equivalent to EQN 2′

[31.8 46.0 1.0

0 18.37 −0.2520 0 5.30

⋮ ⋮ ⋮

0.550.1240.389

]

R3′= R2′ – (18.37/32.93)R3intermediate ,

equivalent to EQN 3′

[31.8 46.0 1.0

0 18.37 −0.2520 32.93 −9.95

⋮ ⋮ ⋮

0.550.124

−0.475]

R3intermediate = R1 – (31.8/9.0)R3,

equivalent to EQN 3intermediate

11

3. Matrix Row-Echelon Method

Alternatively, the matrix can be converted to a row-echelon form in which the diagonal terms in

the matrix are converted to 1’s and the terms below the leading 1’s are converted to 0’s. This is

accomplished by using elementary row operations on columns from left to right to convert the

diagonal terms to 1’s and the terms below the diagonal to 0’s.

Start by writing the absorption equations as an augmented matrix:

In the first column, convert first term in the first row to a 1 and the terms below to 0’s:

In the second column, convert the second term in the second row to a 1 and the term below

it to a 0:

In the third column, convert the third term in the third row to a 1:

In this row-echelon form, CComp3 is given in the third row, CComp3 = 0.0734, or 0.073 g/L. Back-

substitution of the CComp3 value into the second row gives Cβ = 0.00675 – (-0.0137)(0.0734) =

0.00776, or 0.0078 g/L, and back substitution of the CComp3 and Cβ values into the first row gives

Cα = 0.0173 – (0.0314)(0.0734) – (1.447)(0.00776) = 0.00376, or 0.0038 g/L.

B. Solving for Cα, Cβ, and CComp3 With Gauss-Jordan Elimination

The matrix can be converted to a reduced row-echelon form using Gauss-Jordan elimination, in

which the diagonal terms in the matrix are converted to 1’s and all off-diagonal terms are

converted to 0’s. With this technique, back-substitution is eliminated and the concentrations can

be read directly from the matrix.

The matrix is first converted to a row-echelon form using the Row-Echelon Matrix Method

described above. Elementary row operations are then used on columns from left to right to

convert the upper, off-diagonal terms to 0’s. To do this without disturbing the 1’s on the

R1′ = R1/31.8

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

⋮ ⋮ ⋮

0.550.510.29

]

[1 1.447 0.03140 18.37 −0.2520 32.93 −9.95

⋮ ⋮ ⋮

0.01730.124

−0.475]

R2′ = R1 – (31.8/38.1)R2

R3′ = R1 – (31.8/9.0)R3

R2′′ = R2′/18.37

R3′′ = R2′ – (18.37/32.93)R3′

R1

R2 R3

[1 1.447 0.03140 1 −0.01370 0 5.299

⋮ ⋮ ⋮

0.01730.00675

0.339]

[1 1.447 0.03140 1 −0.01370 0 1

⋮ ⋮ ⋮

0.01730.006750.0734

]

R3′′′ = R3′′/5.299

12

diagonal, a row with a 1 on the diagonal is multiplied by a coefficient and subtracted from a row

above it to remove the off diagonal term in that row.

Starting with the row-echelon form from above:

To remove the 1.447 from the second column of the first row while retaining the 1 in the first

column of this row, the second row is multiplied by the coefficient (1.447/1) and subtracted from

the first row.

Similarly, row three is multiplied by the coefficient (-.0137/1) and subtracted from row two, to

make place a zero in the second row third column, and row three is multiplied by (.0516/1) and

subtracted from the first row to remove change the last off-diagonal element to a zero.

The concentrations can now be read directly from the matrix, with Cα = 0.00374, or 0.0037 g/L;

Cβ = 0.00776, or 0.0078 g/L; and CComp3 = 0.0734, or 0.073 g/L.

C. Deriving Concentration Equations Using an Inverse Matrix

The three absorption equations…

A355 = 31.8 Cα + 46.0 Cβ + 1.0 CComp3

A325 = 38.1 Cα + 33.1 Cβ + 1.5 CComp3 The absorption equations

A275 = 9.0 Cα + 3.7 Cβ + 3.1 CComp3

…can be manipulated to derive the three concentration equations:

Cα = –0.516 A355 + 0.0738 A325 – 0.0191 A275

Cβ = 0.0555 A355 – 0.0476 A325 + 0.00510 A275 The concentration equations

CComp3 = 0.0834 A355 – 0.157 A325 + 0.372 A275

[1 1.447 0.03140 1 −0.01370 0 1

⋮ ⋮ ⋮

0.01730.006750.0734

]

[1 0 0.05160 1 −0.01370 0 1

⋮ ⋮ ⋮

0.007530.006750.0734

]

R1

R2 R3

R1′ = R1 – (1.447/1)R2

[1 0 00 1 00 0 1

⋮ ⋮ ⋮

0.003740.007760.0734

]

R1′′ = R1′ – (0.0516/1)R3

R2′′ = R2 – (-0.0137/1)R3

13

Through the use of matrix inversion. Let A be a square matrix, such as the 3 X 3 matrix

representing the absorption equations. Then the inverse matrix A-1 can be defined by:

AA-1 = I = A-1A

Where I is the identity matrix.

In this technique, the three absorption equations…

31.8 Cα + 46.0 Cβ + 1.0 CComp3 = A355

38.1 Cα + 33.1 Cβ + 1.5 CComp3 = A325

9.0 Cα + 3.7 Cβ + 3.1 CComp3 = A275

…are written in matrix form:

or

The matrix equation AX = B is then solved by the simple algebraic steps:

A-1AX = A-1B

And, since A-1A = the identity matrix

X=A-1B

where A-1 is the inverse matrix of the A matrix…

or

…which, written in linear equation form is:

x11 A355 + x12 A325 + x13 A275 = Cα

x21 A355 + x22 A325 + x23 A275 = Cβ The concentration equations

x31 A355 + x32 A325 + x33 A275 = CComp3

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

] [

𝐶𝛼

𝐶𝛽

𝐶3𝑟𝑑

] = [

𝐴355

𝐴325

𝐴275

]

A X = B

[

𝑥11 𝑥12 𝑥13

𝑥21 𝑥22 𝑥23

𝑥31 𝑥32 𝑥33

] [

𝐴355

𝐴355

𝐴275

] = [

𝐶𝛼

𝐶𝛽

𝐶3𝑟𝑑

]

A-1 B = X

14

So, once the inverse matrix is known and each absorbance measured, the concentrations Cα, Cβ,

and CComp3 can calculated by multiplying the inverse matrix by the concentration matrix. The

following are some methods of finding the inverse matrix.

1. Finding the Inverse Matrix Through a System of Linear Equations

One method of finding the inverse matrix is solve AA-1 = I for A-1.

Solve for A-1

Matrix multiplication on the left side of the equal sign gives:

This equates to the following three systems of equations:

Each of these systems of three equations-three unknowns can be solved using the techniques

described above to determine all nine xij terms in the inverse matrix. For example, x11, x21, and

x31 can be found from the first set of equations using the same Gauss-Jordan elimination steps

outlined above. Note that the only difference is that the coefficients on the right hand side of the

augmented matrix are different here.

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

] [

𝑥11 𝑥12 𝑥13

𝑥21 𝑥22 𝑥23

𝑥31 𝑥32 𝑥33

] = [1 0 00 1 00 0 1

]

A A-1 = I

[

31.8𝑥11 + 46.0𝑥21 + 1.0𝑥31 31.8𝑥12 + 46.0𝑥22 + 1.0𝑥32 31.8𝑥13 + 46.0𝑥23 + 1.0𝑥33

38.1𝑥11 + 33.1𝑥21 + 1.5𝑥31 38.1𝑥12 + 33.1𝑥22 + 1.5𝑥32 38.1𝑥13 + 33.1𝑥23 + 1.5𝑥33

9.0𝑥11 + 3.7𝑥21 + 3.1𝑥31 9.0𝑥12 + 3.7𝑥22 + 3.1𝑥32 9.0𝑥13 + 3.7𝑥23 + 3.1𝑥33

] = [1 0 00 1 00 0 1

]

31.8𝑥11 + 46.0𝑥21 + 1.0𝑥31 = 1

38.1𝑥11 + 33.1𝑥21 + 1.5𝑥31 = 0

9.0𝑥11 + 3.7𝑥21 + 3.1𝑥31 = 0

31.8𝑥12 + 46.0𝑥22 + 1.0𝑥32 = 0

38.1𝑥12 + 33.1𝑥22 + 1.5𝑥32 = 1

9.0𝑥12 + 3.7𝑥22 + 3.1𝑥32 = 0

31.8𝑥13 + 46.0𝑥23 + 1.0𝑥33 = 0

38.1𝑥13 + 33.1𝑥23 + 1.5𝑥33 = 0

9.0𝑥13 + 3.7𝑥23 + 3.1𝑥33 = 1

15

Starting with augmented matrix for the first set of equations, and matching the steps listed in the

Row-Echelon method:

Re-labeling the rows to match the notation listed in the Gauss-Jordan method:

Thus x11 = -0.0516, x21 = 0.0555, and x31 = 0.0834. This process can be repeated for the other

two sets of equations to find the other six xij terms to give:

Or Cα = –0.516 A355 + 0.0738 A325 – 0.0191 A275

Cβ = 0.0555 A355 – 0.0476 A325 + 0.00510 A275 The concentration equations

CComp3 = 0.0834 A355 – 0.157 A325 + 0.372 A275

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

⋮ ⋮ ⋮

100

]

[1 1.447 0.03140 18.37 −0.2520 32.93 −9.95

⋮ ⋮ ⋮

0.031411

]

R1′ = R1/31.8

R2′ = R1 – (31.8/38.1)R2

R3′ = R1 – (31.8/9.0)R3

R2′′ = R2′/18.37

R3′′ = R2′ – (18.37/32.93)R3′

R1

R2 R3

[1 1.447 0.03140 1 −0.01370 0 5.299

⋮ ⋮ ⋮

0.03140.05440.442

]

[1 1.447 0.03140 1 −0.01370 0 1

⋮ ⋮ ⋮

0.03140.05440.0834

]

R3′′′ = R3′′/5.299

[1 1.447 0.03140 1 −0.01370 0 1

⋮ ⋮ ⋮

0.03140.05440.0834

]

[1 0 0.05160 1 −0.01370 0 1

⋮ ⋮ ⋮

−0.04730.05440.0834

]

R1

R2 R3

R1′ = R1 – (1.447)R2

[1 0 00 1 00 0 1

⋮ ⋮ ⋮

−0.05160.05550.0834

]

R1′′ = R1′ – (0.0516)R3

R2′′ = R2 – (-0.0137)R3

[−0.0516 0.0738 −0.01910.0555 −0.0476 0.005100.0834 −0.157 0.372

] [

𝐴355

𝐴355

𝐴275

] = [

𝐶𝛼

𝐶𝛽

𝐶3𝑟𝑑

]

16

2. Finding the Inverse Matrix Through Gauss-Jordan Elimination of Adjoined Identity and

Coefficient Matrices

A shorter procedure for finding the inverse matrix is to solve all three systems of equations

simultaneously by adjoining the identity matrix to the coefficient matrix to form [A : I], and then

using Gauss-Jordan elimination to convert [A : I] to [I : A-1]. The steps are identical to those

listed above, but are now also applied to the 3 X 3 identity matrix on the right of the augmented

matrix.

The three concentration equations can now be written directly from this matrix (note that the part

on the right is the inverse matrix):

Cα = –0.516 A355 + 0.0738 A325 – 0.0191 A275

Cβ = 0.0555 A355 – 0.0476 A325 + 0.00510 A275 The concentration equations

CComp3 = 0.0834 A355 – 0.157 A325 + 0.372 A275

[31.8 46.0 1.038.1 33.1 1.59.0 3.7 3.1

⋮ ⋮ ⋮

1 0 00 1 00 0 1

] R1

R2 R3

[1 1.447 0.03140 18.37 −0.2520 32.93 −9.95

⋮ ⋮ ⋮

0.0314 0 01 −0.835 01 0 −3.533

]

R1′ = R1/31.8

R2′ = R1 – (31.8/38.1)R2

R3′ = R1 – (31.8/9.0)R3

[1 1.447 0.03140 1 −0.01370 0 5.299

⋮ ⋮ ⋮

0.0314 0 00.0544 −0.0455 00.442 −0.835 1.971

] R2′′ = R2′/18.37

R3′′ = R2′ – (18.37/32.93)R3′

[1 1.447 0.03140 1 −0.01370 0 1

⋮ ⋮ ⋮

0.0314 0 00.0544 −0.0455 00.0834 −0.157 0.372

] R3′′′ = R3′′/5.299

[1 0 0.05160 1 −0.01370 0 1

⋮ ⋮ ⋮

−0.0473 0.0658 00.0544 −0.0455 00.0834 −0.157 0.372

] R1

iv = R1′′′ – (1.447)R2

[1 0 00 1 00 0 1

⋮ ⋮ ⋮

−0.0516 −0.0738 −0.01910.0555 −0.0476 0.005100.0834 −0.157 0.372

]

R1v = R1

iv – (0.0516)R3′′′

R2v = R2′′ – (-0.0137)R3′′′

17

3. Finding the Inverse Matrix Using a TI Calculator

The inverse matrix can be quickly found using a TI calculator such as the TI-83 or TI-84.

a. Enter the matrix as [[31.8,46.0,1.0][38.1,33.1,1.5][9.0,3.7,3.1]]

b. Hit the X-1 inverse key.

c. Hit the ENTER key.

The inverse matrix will then be displayed.

4. Finding the Inverse Matrix Using Excel

The inverse matrix can be quickly found using Excel.

a. Enter the coefficients of the absorbance equations into a 3 X 3 array of cells.

b. Name the coefficient array “A” by highlighting the array, right clicking in the highlighted

cells, selecting “Name a Range…”, typing A in the Name box, and hitting “OK”. (Of

course, you could name the array anything you want.)

c. Highlight an empty 3 X 3 array of cells into which the inverse matrix will go.

d. While the empty 3 X 3 array of cells is still highlighted, type into the formula bar:

=MINVERSE(A)

e. Hit CTRL SHIFT ENTER to display answers in all of the cells.

f. The inverse matrix will now be displayed in the selected 3 X 3 array of cells:

31.8 46 1

38.1 33.1 1.5

9 3.7 3.1

-0.05156 0.0737856 -0.019071

0.0555703 -0.047586 0.0050997

0.0833633 -0.15742 0.3718601

18

D. Solving for Cα, Cβ, and CComp3 With a TI Calculator

The concentrations can be quickly calculated from the absorbance equations using a TI

calculator, such as a TI-83 or TI-84. Calculators work well using the inverse matrix, calculating

the solution in the form of X = A-1B, where A-1 is the inverse of the 3 X 3 coefficient matrix, B is

the 3 X 1 matrix of absorbances, and X is the 3 X 1 matrix of concentrations.

To calculate the concentrations, key in [[31.8,46.0,1.0][38.1,33.1,1.5][9.0,3.7,3.1]] followed by

the X-1 key followed by *[[.55],[.51],[.29]] and hit ENTER. The calculator will display Cα, Cβ,

and CComp3 .

E. Solving for Cα, Cβ, and CComp3 With Excel

The concentrations can be quickly calculated using Excel. Excel works well using the inverse

matrix, calculating the solution in the form of X = A-1B, where A-1 is the inverse of the 3 X 3

coefficient matrix, B is the 3 X 1 matrix of absorbances, and X is the 3 X 1 matrix of

concentrations.

1. Enter the coefficients of the absorbance equations into a 3 X 3 array of cells.

2. Name the coefficient array “A” by highlighting the array, right clicking in the highlighted

cells, selecting “Name a Range…”, typing A in the Name box, and hitting “OK”. (Of

course, you could name the array anything you want.)

3. Enter the absorbances in a 3 X 1 array of cells

4. Name the absorbance array “B” by highlighting the array, right clicking in the

highlighted cells, selecting “Name a Range…”, typing B in the Name box, and hitting

“OK”. (Of course, you could name the array anything you want.)

5. Highlight an empty 3 X 1 array of cells into which the concentrations will go.

6. While the empty 3 X 1 array of cells is still highlighted, type into the formula bar:

=MMULT(MINVERSE(A),B)

7. Hit CTRL SHIFT ENTER to display answers in all of the cells.

31.8 46 1

38.1 33.1 1.5

9 3.7 3.1

0.55

0.51

0.29

19

8. The concentrations Cα, Cβ, and CComp3 will now be displayed in the selected 3 X 1 array

of cells:

0.0037424

0.0077736

0.0734052