MATH 240

Fri. August 30th, 2019

Chap. 1.2

Basic variable -- variables in a pivot column in reduced echelon form
Free variable -- any non-basic variable

Find the general solution:
(162−5−2−4002−8−13000017)R2+R3→(162−5−2−4002−8010000017)R1+2R3→(162−5010002−8010000017)R2∗12→(162−5010001−405000017)R1−2R2→(160300001−405000017)\begin{aligned} \begin{pmatrix} 1&6&2&-5&-2&-4\\ 0&0&2&-8&-1&3\\ 0&0&0&0&1&7 \end{pmatrix}\\ R_2 + R_3\rightarrow \begin{pmatrix} 1&6&2&-5&-2&-4\\ 0&0&2&-8&0&10\\ 0&0&0&0&1&7 \end{pmatrix}\\ R_1 + 2R_3\rightarrow \begin{pmatrix} 1&6&2&-5&0&10\\ 0&0&2&-8&0&10\\ 0&0&0&0&1&7 \end{pmatrix}\\ R_2 * \frac{1}{2}\rightarrow \begin{pmatrix} 1&6&2&-5&0&10\\ 0&0&1&-4&0&5\\ 0&0&0&0&1&7 \end{pmatrix}\\ R_1 - 2R_2\rightarrow \begin{pmatrix} 1&6&0&3&0&0\\ 0&0&1&-4&0&5\\ 0&0&0&0&1&7 \end{pmatrix} \end{aligned}
{x1=−6x2−3x4x2 is freex3=5+4x4x4 is freex5=7\begin{cases} x_1=-6x_2-3x_4\\ x_2 \ \text{is free}\\ x_3=5+4x_4\\ x_4 \ \text{is free}\\ x_5=7 \end{cases}

A linear system is consistent if and only if the rightmost column of the reduced echelon matrix is not a pivot column.

If consistent: There's only a unique solution if there's no free variables. (Free variables imply infinite solutions)

True or false

  1. "A matrix can be reduced to different reduced echelon forms." (FALSE)
  2. "A matrix can be reduced to different echelon forms." (TRUE)
  3. "Row reduction only applies to augmented matrices." (FALSE)
  4. "Basic variables correspond to pivot columns in the augmented matrix." (FALSE (only true for coefficient matrices))
  5. "If one row of the echelon form of an augmented matrix is [00050]\begin{bmatrix} 0&0&0&5&0 \end{bmatrix}, then the solutions are consistent." (FALSE (can't determine without knowing other rows))
  6. "Whenever a system has free variables, the solution set is infinite." (FALSE (only true for consistent systems))
  7. "If two systems of linear equations have the same solution set, then the associated row reduced echelon forms of the augmented matrices are the same." (FALSE)