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Linear Algebra and Geometry

Systems of Linear Equations and Gaussian Elimination

A system of linear equations in matrix form is written as Ax = b, where A is the coefficient matrix, x is the variable vector, and b is the constant vec…

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Linear Systems & Gaussian Form

A system of linear equations in matrix form is written as Ax=bA\vec{x} = \vec{b}, where AA is the coefficient matrix, x\vec{x} is the variable vector, and b\vec{b} is the constant vector.

Gaussian elimination solves this by reducing the augmented matrix [Ab][A|\vec{b}] into row echelon form using three safe operations:

OperationNotationRule
SwapRiRjR_i \leftrightarrow R_jExchange rows
ScaleRiαRiR_i \to \alpha R_iα0\alpha \neq 0
AddRiRi+αRjR_i \to R_i + \alpha R_jAdd multiple

Example: Solve a 2×22\times 2 system: (215411)R22R1(215039)\begin{pmatrix} 2 & 1 & | & 5 \\ 4 & -1 & | & 1 \end{pmatrix} \xrightarrow{R_2 - 2R_1} \begin{pmatrix} 2 & 1 & | & 5 \\ 0 & -3 & | & -9 \end{pmatrix}

Back-substitution yields y=3y = 3 and x=1x = 1. Row operations preserve the solution set, but yield a different matrix telling the same story.

Solution Types & Ranks

Always check the rank of both AA and the augmented matrix [Ab][A|\vec{b}] before interpreting your final system outcome.

Solution TypeRank ConditionMeaning
Uniquerk(A)=rk([Ab])=n\mathrm{rk}(A) = \mathrm{rk}([A|\vec{b}]) = nExact fit
Infiniterk(A)=rk([Ab])<n\mathrm{rk}(A) = \mathrm{rk}([A|\vec{b}]) < nFree variables
Nonerk(A)<rk([Ab])\mathrm{rk}(A) < \mathrm{rk}([A|\vec{b}])Contradiction

Common pitfall: A resulting row of the form 0  0    c0\;0\;|\;c where c0c \neq 0 instantly means no solution. Stop immediately when this contradiction appears.

Placeholder: Systems of Linear Equations and Gaussian Elimination

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Linear Algebra and Geometry