The system \begin{cases} x + y = 3, \\ x - y = -1, \\ 2x - y = 4 \end{cases} \\ Option 1: has no solution \\ Option 2: has multiple solution \\ Option 3: has a unique solution \\ Option 4: None of these
Added by Ashlee K.
Close
Step 1
First, let's rewrite the system in matrix form: X - Zx = J Show more…
Show all steps
Your feedback will help us improve your experience
Teresa Fuston and 64 other Algebra 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
Consider the system $\left\{\begin{array}{l}k x+y+z=1 \\ x+k y+z=1 \\ x+y+k z=1\end{array}\right.$ Use determinants to find those values of $k$ for which the system has (a) a unique solution, (b) more than one solution, (c) no solution.
The system has no solution. The system has unique solution: (x,y) = ( , ) The system has infinitely many solutions. They must satisfy the following equation: y = The system has no solution. The system has unique solution: (x,y) = ( , ) The system has infinitely many solutions. They must satisfy the following equation: y =
Supreeta N.
The system of equations $x-k y-z=0, k x-y-z=0$ and $x+y-z=0$ has non trivial solutions. Then the possible values of $k$ are (A) $\pm 1$ (B) $\pm 2$ (C) 0 (D) $\pm 4$
Engineering Mathematics
Linear Algebra
Recommended Textbooks
Elementary and Intermediate Algebra
Algebra and Trigonometry
Watch the video solution with this free unlock.
EMAIL
PASSWORD