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01 System of Linear Equations

Elementary Row Operations

\[ A = \begin{bmatrix} 1 & 2 & -1 \\ -9 & 6 & 4 \\7 & 3 & -1 \end{bmatrix}_{3 \times 3} \]
  • Any 2 rows can be interchanged

    \(R_1 \iff R_2\) - Any row can be multiplied/divided by any number other than 0

    \(R_1 \to 2R_1\) - Any row can be added/subtracted to any row

    \(R_1 \to R_1 \pm 2 R_2\)

REF

Reduced Echelon Form

Upper \(\triangle\)r matrix

  • 1st non-zero elment in a row should be 1

    (called as leading one) - Leading one should occur to the right side of previous rows’ leading one(s) - If there is any zero row, it should be the last row otherwise, we need to interchange rows to ensure this rule

example

\[ \begin{bmatrix} 1 & 4 & 5 & 3 \\ 0 & 1 & 2 & 8 \\0 & 0 & 1 & 5 \end{bmatrix} \quad \begin{bmatrix} 1 & 4 & 3 & 5 \\ 0 & 1 & 8 & 2 \\0 & 0 & 0 & 1 \end{bmatrix} \]

RREF

diagonal matrix

is the REF matrix where the elements of the columns of the leading ones (other than itself) are 0.

\[ \begin{bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 8\\0 & 0 & 1 & 5 \end{bmatrix} \quad \begin{bmatrix} 1 & 0 & 5 & 0\\ 0 & 1 & 8 & 0\\0 & 0 & 0 & 1 \end{bmatrix} \]

Rank

no of non-zero rows of a matrix in REF/RREF

Gauss Methods

Method Form
Gauss Elimination REF
Gauss Jordan RREF
  1. Write equation in matrix form \(AX = B\), where
    • \(A\) is coefficients matrix
    • \(B\) is constant matrix
    • \(X\) is variable matrix

Converted augmented matrix = \([A | B]\) into REF

  1. Cases

    \(n\) is the number of unknown variables

Rank(A\vert B)
\(\ne\) rank(A) no solutions
\(=\) rank(A) \(= n\) unique solutions
\(=\) rank(A) \(< n\) infinite solutions
  1. Back Substitution

    Degree of freedom = no of vars - no of equations

Homogeneous Linear System

There will always be a solution.

If there is unique solution, it is always all 0s. This is called as trivial solution.

Inverse of matrix

If \(A\) and \(B\) are 2 non-singular matrices such that \(|A| \ne 0\), then \(A^{-1} = B \iff A\cdot B = I\)

\(I\) is identity matrix

\[ \begin{aligned} I_{2 \times 2} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ I_{3 \times 3} &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} \end{aligned} \]

To find inverse

  • use row transformations to convert \([A:I] \to [I:B]\)
  • then \(B = A^{-1}\)

If \(A\) is singular, inverse does not exist

Last Updated: 2024-01-24 ; Contributors: AhmedThahir

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