Algebra question

amathproblemthatneedsolve

amathproblemthatneedsolve

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This is a linear systems question:

Equation 1: 3x+y - 2z = -7

Equation 2: -az = x+y+6

To make the 3rd equation use the digits 2,2,1,9 they will be the coefficients of x,y,z, and constant term in that order, given the equation is in form ax+by+cz=d

1. write the 3rd equation and sole for the set of three equations when a= -3

Writing the 3rd one is this right? 2x+2y+z=9

solving the set of three equations. :

3x+y - 2z = -7

x+y-3z=-6

2x+2y+z=9

x,y,z = -2,5,3




2. find the values of a, for which the system is consistent. Give algebraic reasoning for your answer and interpret the solution geometrically


Not sure?


3. new set of equations where you use equation 1 and 2 again but equation 3 is made by timesing equation 1 by 3

3x+y - 2z = -7

x+y-3z=-6

9x+3y-6z=-21

working:
x,y,z= 0,-9,-1


4. new set of equations where you use equation 1 and 2 again but equation 3 is made by changing the constant -7 to 12

3x+y - 2z = -7

x+y-3z=-6

3x+y - 2z = 12

working:

x,y,z= I get nothing
 
2) Solve without replacing "a" and see where it leads.
 
2) Solve the system using a for a. That is solve this system of equations
3x+y - 2z = -7
x+y+az=-6
2x+2y+z=9

3) You say "working" but you do not show your work. Your solution is not correct (it works for all three eqs but so do other combinations) so please show your work so we can where you made a mistake.

4) What do you mean you got nothing. Does not mean you could not solve the problem. Nothing is not a solution to a system of equations. Please state what you mean, using mathematical terminology or show us your work.
 
Jomo said:
2) Solve the system using a for a. That is solve this system of equations
3x+y - 2z = -7
x+y+az=-6
2x+2y+z=9
Click to expand...
If you can find an \(a\) that makes this matrix non-singular:
\(A=\left( {\begin{array}{*{20}{c}} 3&1&{ - 2} \\ 1&1&a \\ 2&2&1 \end{array}} \right)\)
Find the value(s) of \(a)\) that must be excluded so the determinate \(|A|\ne 0\).
"SEE HERE"
 
From the the way the question is posed I just feel that this did not come from a Linear Algebra course but rather an Algebra course. In an algebra class students do not learn about non-singular matrices.
 
Jomo said:
2) Solve the system using a for a. That is solve this system of equations
3x+y - 2z = -7
x+y+az=-6
2x+2y+z=9

3) You say "working" but you do not show your work. Your solution is not correct (it works for all three eqs but so do other combinations) so please show your work so we can where you made a mistake.

4) What do you mean you got nothing. Does not mean you could not solve the problem. Nothing is not a solution to a system of equations. Please state what you mean, using mathematical terminology or show us your work.
Click to expand...

for 1) Do they want me to solve it so I get a=-3? if so When I solve where a=-3 is.

a=-3
x+y-3z=-6
3x+y-2z=-7
2x+2y+z= 9

x+y-3z=-6
-2y+7z=11
7z=21
x=-2, y=5, z=3

2)

x=-(41+23 a)/(4a -2), y=(41a+53)/(4a-2), z=-21/(2a-1
Now if we look at the expressions for x, y and z we see that all of them have 2a-1 (or 4a-2 ) in the denominator. So, the system has a unique solution for a≠1/2.
Now the question is: what happens when a=1/2. In that case, the system has no solutions.

3) You're meant to continue using a=-3, so I get look at the link is this right?

a=-3, 3x+y-2z=-7, x+y-3z=-6 , 9x+3y-6z=-21 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
www.wolframalpha.com www.wolframalpha.com


4) You're meant to continue using a=-3, And There are no solutions that exist?
 
#2: If you did the work correctly-why do you refuse to show your work? - then you are correct that there are no solutions when a=1/2.

#3) The problem says to use equations 1 and 2 which does have an a in it. I guess that the author did not want you to use -3 for a.
If your equations do not have an a in it (since you replaced a with -3) then why at WA do you start off by saying a=-3. very strange.

#4)new set of equations where you use equation 1 and 2 again but equation 3 is made by changing the constant -7 to 12.
The problem I see here is that equation 3 is 2x+2y+z=9 and there is no constant -7 to change to 12
 
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