Question

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,

Forms of linear equations
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asked 2021-05-23

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

Answers (1)

2021-05-24

The system is consistent and has infinitely many solutions.
Parameter: \(\displaystyle{x}{3}∈{R}\)
Row 2: \(\displaystyle{x}{2}+{3}{x}{3}={1}\to{x}{2}={1}-{3}{x}{3}\)
Row 1: \(\displaystyle{x}{1}-{2}{x}{3}={6}\to{x}{1}={6}+{2}{x}{3}\)

 Infinitely many solutions,

\(\begin{cases}x_1=6+2x_3\\x_2=1-3x_3\end{cases}, x_3 ∈ R\)

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