Solve the system of linear equations using Gaussian elimination with back-substitution. $$\begin{array}{r} x-2 y+3 z=1 \\ -2 x+7 y-9 z=4 \\ x+z=9 \end{array}$$
Added by Eugene P.
Step 1
$$\left[\begin{array}{ccc|c} 1 & -2 & 3 & 1 \\ -2 & 7 & -9 & 4 \\ 1 & 0 & 1 & 9 \end{array}\right]$$ Show more…
Show all steps
Close
Your feedback will help us improve your experience
Jerelyn Nevil and 50 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Solve the system of linear equations using Gaussian elimination with back-substitution. $$\begin{array}{r}x-2 y+3 z=1 \\-2 x+7 y-9 z=4 \\x+z=9\end{array}$$
Systems of Linear Equations and Inequalities
Systems of Linear Equations and Matrices
Solve the system of linear equations using Gaussian elimination with back-substitution. $$\begin{array}{r}2 x+5 y=9 \\x+2 y-z=3 \\-3 x-4 y+7 z=1\end{array}$$
Solve the system of linear equations using Gaussian elimination with back-substitution. $$\begin{array}{rr}2 y+z= & 3 \\4 x-z= & -3 \\7 x-3 y-3 z= & 2 \\x-y-z= & -2\end{array}$$
Recommended Textbooks
Precalculus with Limits
Precalculus
Watch the video solution with this free unlock.
EMAIL
PASSWORD