ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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The rank and solution behaviour of a parameterised matrix change at one exceptional parameter

Example

For $A_t=\begin{pmatrix}1&1\\1&t\end{pmatrix},\qquad b=\begin{pmatrix}2\\2\end{pmatrix},$ the real system $A_tx=b$ has rank $2$ and the unique solution $(2,0)$ when $t\ne1$ , while at $t=1$ it has rank $1$ and the affine line of solutions $(2-s,s)$ .

Facts & Assumptions

Given: A real parameter $t$ and the displayed system.

[L1]

Matrix rank is the number of pivots (Row rank equals column rank, and both equal the number of pivots).

[L2]

RREF gives consistency and the free-variable parametrisation (Reduced row echelon form detects consistency and parametrises every solution by the nonpivot variables).

[L3]

The real numbers form a field (The reals form a field).

Verification

technique · cases

1.1

Subtract row $0$ from row $1$ without dividing by the parameter; the augmented matrix becomes $\begin{pmatrix}1&1&2\\0&t-1&0\end{pmatrix}$ .

L1L2L3cases

2.1

If $t\ne1$ , the second row has a pivot, so the rank is $2$ , $y=0$ , and $x=2$ .

step 1.1assume-case generic

2.2

If $t=1$ , the second row is zero, so the rank is $1$ and $y=s$ is free with $x=2-s$ .

step 1.1assume-case exceptional

3.1

The alternatives $t=1$ and $t\ne1$ are exhaustive and give the stated solution behaviour.

step 2.1step 2.2cases-exhaustive∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Hefferon, Linear Algebra, 4th ed., Ch. One, exercises after §III.2 (standard reference, not scraped)