Matrix Inverses and Linear Systems

A market maker needs to find hedge ratios for three correlated instruments so the combined position has zero net exposure to three separate risk factors. That's three equations in three unknowns — solvable by hand for small cases, but tedious and error-prone by substitution once the system grows. The matrix inverse turns this into one clean operation: write the system as a matrix equation, apply the inverse, and read off every unknown at once.

An analogy: undoing a mixing recipe

Imagine a machine that takes two input amounts and mixes them into two output amounts according to a fixed recipe — say, output 1 is always "2 times input A plus 1 times input B," and output 2 is "1 times input A plus 3 times input B." If you're handed the outputs and asked to figure out what inputs produced them, you need the "un-mixing" recipe — the exact reverse operation. The matrix inverse is that reverse recipe: applied to the machine's matrix, it undoes whatever the original matrix did, turning outputs back into the unique inputs that produced them (as long as the original recipe didn't destroy information by collapsing two different inputs onto the same output).

The mathematics, one symbol at a time

For a square matrix $A$, its inverse $A^{-1}$ is the unique matrix satisfying

$$A A^{-1} = A^{-1} A = I,$$

where $I$ is the identity matrix (the "do nothing" transformation). In words: applying $A$ then $A^{-1}$, in either order, gets you back exactly where you started. The inverse exists if and only if $\det(A) \neq 0$ (see Determinants), equivalently if $A$'s columns are linearly independent. Given a linear system $Ax = b$ — a matrix $A$ of known coefficients, a vector $b$ of known targets, and an unknown vector $x$ to solve for — the solution is

$$x = A^{-1} b,$$

read as: multiplying both sides of $Ax=b$ by $A^{-1}$ cancels the $A$ on the left, leaving $x$ isolated and solved for directly, all unknowns at once. In practice, computers rarely form $A^{-1}$ explicitly — they solve the system via decomposition methods like LU (see LU Decomposition) which are faster and more numerically stable — but the formula $x = A^{-1}b$ captures the idea cleanly.

Worked example 1: a 2×2 hedge ratio system by hand

Solve $\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 8 \\ 13 \end{pmatrix}$. First, the inverse of a $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Here $\det = (2)(3)-(1)(1)=5$, so $A^{-1} = \frac{1}{5}\begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}$. Then $x = A^{-1}b = \frac{1}{5}\begin{pmatrix} 3(8) - 1(13) \\ -1(8) + 2(13) \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 11 \\ 18 \end{pmatrix} = \begin{pmatrix} 2.2 \\ 3.6 \end{pmatrix}$. Check: $2(2.2)+1(3.6) = 4.4+3.6=8$ ✓, and $1(2.2)+3(3.6)=2.2+10.8=13$ ✓.

Worked example 2: solving for hedge notionals

A trader wants a position long $0 exposure to two risk factors using two hedge instruments with sensitivities: instrument 1 has factor sensitivities $(1.2, 0.5)$ per $1, instrument 2 has $(0.3, 1.8)$ per $1, and the existing book has raw exposures $(6.0, 4.5)$ that need to be exactly offset. Solve $\begin{pmatrix} 1.2 & 0.3 \\ 0.5 & 1.8 \end{pmatrix}\begin{pmatrix} n_1 \\ n_2\end{pmatrix} = \begin{pmatrix} 6.0 \\ 4.5 \end{pmatrix}$: determinant $= 1.2(1.8)-0.3(0.5) = 2.16-0.15=2.01$; $n_1 = (1.8(6.0)-0.3(4.5))/2.01 = (10.8-1.35)/2.01 \approx 4.70$, $n_2 = (1.2(4.5)-0.5(6.0))/2.01 = (5.4-3.0)/2.01 \approx 1.19$. Buying about $4.70 of instrument 1 and $1.19 of instrument 2 exactly neutralizes both factor exposures at once.

Drag the transform above, note the transformed shape, then imagine reading it backward: the inverse matrix is exactly the transform that maps the distorted shape back to the original unit square — it exists as long as the shape hasn't been flattened onto a line.

What this means in practice

Solving $Ax=b$ via a matrix inverse (or, better, a stable decomposition) is the standard way to compute hedge ratios, factor-neutral portfolio weights, and the coefficients in ordinary least squares regression, where $x = (X^TX)^{-1}X^Ty$. When $A$ isn't square or isn't invertible — more equations than unknowns, or redundant instruments — the closely related pseudoinverse (see The Moore-Penrose Pseudoinverse) provides the best-fit generalization.