Topic: How do I find values for a so that the system has one solution, no solutions, and infinite solutions?**Question:**
Hello! I just started linear algebra and I am a bit lost. I have a homework problem which says "Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions." The system is:
x + 2y + z = 2
2x - 2y + 3z = 1
x + 2y - (a^2 - 3)z = a
Any help that you could give me would be most appreciated. I have been stuck on this problem for hours and I get the feeling I am overcomplicating it. Thanks!

June 19, 2019 / By Wisdom

assume you have written it out as a matrix and are looking at that. in this case you can think of each equation as a plane. So you will either have planes that intersect in a line (infinitely many answers), planes that intersect in a point (one answer) or planes that don't all intersect. So for the no solution case, you see that the first equation and the third equation share the same co-efficients for X and Y, but not Z. Solve -(a^2-3) = 1 for a to get rad(2). (square root 2). You would have x/y/z co-efficients the same for first and third equation, but in the first case, they equal 2 and in the last case they equal rad(2). These are inconsistent ... so these planes have no intersection and no solution. Same values for X Y Z in the same equation can't equal different things. to see a specific solution ... you are just looking for a consistent answer (X,Y,Z) that solves all three equations. Just solve this as you normally would with Gauss Jordan elimination, and you z = (a -2)/(-a2 +2). Then y = 1//2 + 1/6z. and x = 1 - 4/3z. (the value of z). At any rate, if you pick any value for "a" now, say a=0, you can get x = 7/3, y = 1/3, z = -1... which will solve all three equations. For a=0, you have a solution for all three equations. This solution works for any "a" except a = rad(2) as you see the denominator of Z with a = rad(2) = 0. To get infinitely many solutions, the intersection of the 3 planes must be a line. In this case, since you have all the values of A covered .. there is no solution that will give you a line.

👍 180 | 👎 6

Did you like the answer? Well first and foremost we need to find a way to make it not so expensive. I mean it is crazy there are all these children in orphanges and so many people who want children but, the cost is so high it is just not realistic for most people. I believe it is around 40,000. Why should it cost so much when the child really needs a home. Also making the need for Adoptive parents for over seas more known. Also find some way to make it less confusing on how to go about it. Calling an agency to adopt over seas is also scary because, you never know if it is a ligit company or not.

That's stupid. Americans should be adopting American babies. Let China, Nigeria, and the rest of them figure out their own problems.

Add eqns 1 and 2 3x + 4z = 3 Add eqns 2 and 3 3x + (3 + a^2 - 3)z = 1+a or 3x + a^2z = 1+a Subtract these two results 4z - a^2z = 3 - 1 - a (4-a^2)z = 2 - a If 4-a^2 = 0 (which is a = 2 and - 2) we have either: (a = 2) 0 = 0 (which is ALWAYS true) so we have inf solutions (a=-2) 0 = 4 (which is NEVER true) so we have no solutions If a is anything other than -2 and 2, we have 1 solution

👍 70 | 👎 -2

thus row reduce to a triangular system....now choose " a " so the last line has a solution { 1 } , choose " a " so the last line makes no sense { no } , finally choose a so the last line is all 0's...{ infinite }...one can see if a² = 2 then 1st and last cannot both hold

👍 65 | 👎 -10

you cant come up with one. the first wireless technology regulated and controlled the people, took a step ahead and built this massive framework in a holographic matrix. the problem is obviously technological advancement but you cant expect them to stop. theyll afflict me unto death with as many technologies as they have while occupying my queen

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