Each variable is represented by a leading 1, so there are no parameters needed.

There are no impossible equations , such as (0=1).

The system is consistent and has exactly one solution.

(No parameters) Interpreting row by row as equations.

d2saint0

Answered 2021-09-15
Author has **28283** answers

Each variable is represented by a leading 1, so there are no parameters needed.

There are no impossible equations , such as (0=1).

The system is consistent and has exactly one solution.

(No parameters) Interpreting row by row as equations.

asked 2021-09-25

The reduced row echelon form of the augmented matrix of a system of linear equations is given. Tell whether the system has one solution, no solution, or infinitely many solutions. Write the solutions or, if there is no solution, say the system is inconsistent.

\(\begin{bmatrix}1&0&-2&|&6\\0&1&3&|&1 \end{bmatrix}\)

asked 2021-05-03

asked 2021-06-26

Determine if the statement is true or false, and justify your answer. (a) Different sequence s of row operations can lead to different reduced echelon forms for the same matrix. (b) If a linear system has four equations and seven variables, then it must have infinitely many solutions.

asked 2021-09-19

asked 2021-09-15

asked 2021-10-24

The given set of equations has no solution or infinitely many solutions.

Given: \(\displaystyle{6}{x}{y}\leq{24}\ {\quad\text{and}\quad}\ {6}{x}-{y}{>}{24}\) are the given set of inequalities.

Given: \(\displaystyle{6}{x}{y}\leq{24}\ {\quad\text{and}\quad}\ {6}{x}-{y}{>}{24}\) are the given set of inequalities.

asked 2021-06-19

The reduced row echelon form of the augmented matrix of a system of linear equations is given. Tell whether the system has one solution, no solution, or infinitely many solutions. Write the solutions or, if there is no solution, say the system is inconsistent.

\(\begin{bmatrix}1 & 0&0&|&-1 \\0 & 1&0&|&3\\0 &0 &1&|&4\\0&0&0&|&0\end{bmatrix}\)