3.2 Matrix Algebra

Section 3.1 left the laboratory hanging. The spectrophotometer measured absorbances \mathbf{A} = (0.78, 0.74) for a two-dye mixture; Beer’s law says E\mathbf{c} = \mathbf{A} with the absorptivity matrix E known; and the numbers the chemist is paid for — the concentrations \mathbf{c} — sit locked inside the product. In scalar arithmetic the analogous jam, 5c = 0.78, is escaped by dividing: multiply both sides by the number \tfrac15 that undoes multiplication by 5. This section builds the matrix version of that escape. It requires knowing how to multiply matrices by matrices, which matrix plays the role of 1, and which matrix plays the role of \tfrac15 — and, unlike the scalar story, each of these holds a surprise.

Multiplying grids

Matrices earn a product through composition: doing one linear operation after another. Suppose a two-step process — first raw feeds blend into intermediate streams (a matrix B acting on the feed vector), then intermediates convert into products (a matrix A acting on the result). The overall feeds-to-products operation sends \mathbf{v} to A(B\mathbf{v}), and the question is whether a single matrix performs the combined step. It does, and its recipe follows from the column picture of Theorem 3.1.3: to know any matrix, it is enough to know what it does to each basis direction — its columns.

Definition 3.2.1 (Matrix product). For A (m \times n) and B (n \times p), the product AB is the m \times p matrix whose j-th column is A applied to the j-th column of B. Entry by entry, this works out to (AB)_{ij} = \text{(row } i \text{ of } A) \cdot \text{(column } j \text{ of } B) = \sum_{k=1}^{n} a_{ik}\, b_{kj}, row of the left factor against column of the right. AB then satisfies (AB)\mathbf{v} = A(B\mathbf{v}) for every vector — it is the composite operation.

The inner dimensions must match (n with n), and the product inherits the outer ones — the same contract logic as §3.1, chained. But the working surprise is this: order matters. Compute both orders for a small pair: \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}. AB \neq BA, and not by accident of these numbers: composition of operations is genuinely order-sensitive — protonate then heat is not heat then protonate, rotate then reflect is not reflect then rotate (§3.3 makes that one visual). Everything you know about scalar algebra transfers to matrices except commutativity, and every manipulation in this chapter must be performed with left/right discipline. Associativity, mercifully, survives: (AB)C = A(BC) — grouping is free even though order is not (Exercise 10 of §3.1 met a first instance).

The matrix that does nothing

Scalar 1 is the number that multiplies without changing. Its matrix counterpart must send every vector to itself, and the column picture builds it instantly: its columns must be where the basis vectors land — which is, unchanged, the basis vectors themselves.

Definition 3.2.2 (Identity). The n \times n identity matrix I has ones on the diagonal and zeros elsewhere. It satisfies I\mathbf{v} = \mathbf{v} for every vector and AI = IA = A for every compatible matrix.

Undoing: the inverse

Now the main event. To free \mathbf{c} from E\mathbf{c} = \mathbf{A} we want a matrix that, multiplied against E, leaves the identity — the matrix analogue of \tfrac15.

Definition 3.2.3 (Inverse). A square matrix A is invertible if there exists a matrix A^{-1} with A^{-1} A = A A^{-1} = I. Then A\mathbf{x} = \mathbf{b} is solved by one multiplication: \mathbf{x} = A^{-1}\mathbf{b} — apply A^{-1} to the left of both sides and let A^{-1}A collapse to I.

For 2 \times 2 matrices the inverse has a closed formula, worth knowing cold both for daily use and for the number it forces into view.

Theorem 3.2.4 (The 2×2 inverse). For A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, if ad - bc \neq 0 then A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} — swap the diagonal, negate the off-diagonal, divide by ad - bc. If ad - bc = 0, no inverse exists.

Proof. Multiply and watch the cancellation: \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} ad - bc & -ab + ba \\ cd - dc & -cb + da \end{pmatrix} = (ad - bc)\, I . Dividing by the scalar ad - bc (legitimate exactly when it is nonzero) gives A A^{-1} = I; the product in the other order works out identically. For the failure case: if ad - bc = 0 the rows of A are proportional, so every product A\mathbf{x} has proportional entries too — the outputs of A fill only a line, not the plane, and no matrix can undo a collapse that loses information. (What that loss means, and how ad-bc measures it, is §3.4’s whole subject.)

The combination ad - bc was not invited — it emerged, uninvited, as the exact quantity whose vanishing kills invertibility. Hold onto it. For now, the formula pays the section’s debt.

Example 3.2.5 (The mixture, resolved). At two wavelengths, the dye pair of §3.1 has E = \begin{pmatrix} 15000 & 3000 \\ 2000 & 11000 \end{pmatrix} M⁻¹cm⁻¹ (\ell = 1 cm), and the instrument reads \mathbf{A} = (0.78, 0.74). Find the concentrations.

Setting it up. \mathbf{c} = E^{-1}\mathbf{A}. First the crucial scalar: ad - bc = (15000)(11000) - (3000)(2000) = 1.65 \times 10^8 - 6 \times 10^6 = 1.59 \times 10^8 — nonzero, so the system is solvable and the dyes are spectroscopically distinguishable.

Solution. By Theorem 3.2.4, E^{-1} = \frac{1}{1.59 \times 10^8} \begin{pmatrix} 11000 & -3000 \\ -2000 & 15000 \end{pmatrix}, c_X = \frac{11000(0.78) - 3000(0.74)}{1.59 \times 10^8} = \frac{8580 - 2220}{1.59 \times 10^8} = \frac{6360}{1.59 \times 10^8} = 4.0 \times 10^{-5}\ \text{M}, c_Y = \frac{-2000(0.78) + 15000(0.74)}{1.59 \times 10^8} = \frac{-1560 + 11100}{1.59 \times 10^8} = \frac{9540}{1.59 \times 10^8} = 6.0 \times 10^{-5}\ \text{M}.

Check. Round trip: feed \mathbf{c} forward through E — 15000(4.0\times10^{-5}) + 3000(6.0\times10^{-5}) = 0.60 + 0.18 = 0.78 ✓ and 2000(4.0\times10^{-5}) + 11000(6.0\times10^{-5}) = 0.08 + 0.66 = 0.74 ✓; the inverse’s only job is to make the forward product reproduce the data, and it does, exactly. Units: (M⁻¹cm⁻¹)⁻¹ × (dimensionless absorbance) gives M·cm per cm of path — molar ✓. Magnitudes: tens of micromolar, the natural range for dyes with \varepsilon \sim 10^4 showing A \sim 1 ✓. And these are the very concentrations §3.1 used to generate those absorbances — the loop closes.

When undoing fails — and when it barely succeeds

Theorem 3.2.4’s exclusion, ad - bc = 0, is not a mathematical technicality; it is a spectroscopic statement. The quantity vanishes exactly when the columns of E are proportional — when dye Y’s absorptivity pattern across the wavelengths is a scalar multiple of dye X’s. Two dyes with proportional spectra are indistinguishable at those wavelengths no matter how precise the instrument: a large amount of a weak absorber and a small amount of a strong one produce identical readings, information is genuinely absent, and no algebra can conjure it. The practical corollary bites even before exact proportionality: if the columns are nearly proportional, ad - bc is small, the \tfrac{1}{ad-bc} out front is huge, and tiny absorbance noise is amplified into wild concentration error. Analytical chemists know the fix instinctively — measure at wavelengths where the dyes differ most — and they are, in this chapter’s language, choosing wavelengths to push E’s columns away from proportionality. Section 3.4 will name the quantity, generalize it beyond 2 \times 2, and reveal it as a volume.

Remarks and cautions

Never write matrix division. \mathbf{b}/A is meaningless because it hides the one thing that matters: which side. Undoing is always explicit multiplication by A^{-1}, on a stated side — A^{-1}\mathbf{b} for our systems — and on the same side of both halves of an equation.

Inverses reverse order. (AB)^{-1} = B^{-1} A^{-1} — socks-then-shoes is undone shoes-then-socks. Expanding (AB)^{-1} as A^{-1}B^{-1} is the non-commutativity error in its most seductive costume (Exercise 8 makes you prove the correct version, which takes two lines).

Not every square matrix has an inverse. The scalar world prepares you badly here: among numbers only 0 lacks a reciprocal, but singular matrices — whole families of them — are everywhere, and chemistry manufactures them whenever an experiment’s measurements are redundant. Check ad - bc before dividing by it, the same reflex as §2.3’s clamp-before-arccos.

Small ad - bc is almost as bad as zero. An inverse that exists on paper can be worthless in the lab if it multiplies measurement noise a thousandfold. When concentrations from a perfectly correct formula come out absurd, inspect the matrix, not the algebra: the instrument may be answering a question the wavelength choice made nearly unanswerable.

Summary

Matrix multiplication is composition — column j of AB is A applied to column j of B, equivalently row-into-column dot products — and it obeys all of scalar algebra’s rules except commutativity, which fails structurally because sequential operations are order-sensitive. The identity I plays the role of 1, and an invertible matrix has an A^{-1} that collapses against it, solving A\mathbf{x} = \mathbf{b} in one left-multiplication; for 2 \times 2 the inverse is the swap–negate–divide formula of Theorem 3.2.4, whose divisor ad - bc decides everything: nonzero means solvable (the Beer’s-law mixture yielded its concentrations to exactly this formula, with a round-trip check), zero means information was never there (proportional spectra), and small means the answer exists but amplifies noise. The uninvited quantity ad-bc now has two jobs — invertibility test and noise amplifier — and §3.4 will show both are shadows of one geometric fact.

Exercises

  1. For A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} and B = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix}, compute AB and BA, and confirm they differ. Which entries agree, and is that a coincidence you can explain?

  2. Verify by multiplication that \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}, and use it to solve 3x + y = 7, 5x + 2y = 12.

  3. Compute the inverses, or declare them nonexistent with the reason: \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. For the last: what does that matrix do to a vector, and why is it its own inverse?

  4. A mixture of nitrate and nitrite is measured at 220 and 355 nm: E = \begin{pmatrix} 9000 & 5000 \\ 100 & 23 \end{pmatrix} M⁻¹cm⁻¹… except a units slip has made the second row implausible. Ignore chemistry and solve anyway for \mathbf{A} = (0.55, 0.0041); then recompute with the corrected E_{22} = 2300 and \mathbf{A} = (0.55, 0.41). Which inverse was more noise-sensitive, and how could you have predicted that from ad - bc alone?

  5. Show that diagonal matrices commute with each other, and exhibit a diagonal matrix and a non-diagonal one that do not commute. (Diagonal-with-diagonal is the special case scalar intuition secretly relies on.)

  6. Two reaction steps act on a stream vector: dilution D = \begin{pmatrix} \tfrac12 & 0 \\ 0 & \tfrac12 \end{pmatrix} and a partial conversion C = \begin{pmatrix} 0.9 & 0 \\ 0.1 & 1 \end{pmatrix} (10% of species 1 becomes species 2). Compute CD and DC. They agree — reconcile this with non-commutativity in one sentence (what is special about D?).

  7. ★ Prove that an inverse, when it exists, is unique: if BA = I and AC = I, insert cleverly into B(AC) and use associativity to force B = C. (This is why “the” inverse is grammatical.)

  8. ★ Prove (AB)^{-1} = B^{-1}A^{-1} by directly checking that B^{-1}A^{-1} does the required job against AB — both orders. Where exactly would A^{-1}B^{-1} fail?

  9. ★ For the dyes of Example 3.2.5, suppose absorbance noise of \pm 0.01 afflicts A_1. Propagate it: recompute c_X with A_1 = 0.79 and report the percent change. Repeat the whole exercise for a “bad” wavelength pair with E' = \begin{pmatrix} 15000 & 3000 \\ 14000 & 2900 \end{pmatrix} (near-proportional rows). What ratio of percent errors do you find, and which single scalar foretold it?

  10. ★★ Derive Theorem 3.2.4 rather than verifying it: write A^{-1} = \begin{pmatrix} p & q \\ r & s \end{pmatrix}, expand AA^{-1} = I into four scalar equations, and solve them by elimination. Note where the derivation forces you to divide by ad - bc — failure is not at the end but woven through.

Practice until it sticks