Linear algebra question find a b c to get infinite number of solutions

[sozener1]

4. Find the values of \(\displaystyle a, \, b,\, c\, \in\, \mathbb{R}\) for which the equations \(\displaystyle x\, +\, y\, +\, 2z\, =\, a,\) \(\displaystyle x\, +\, z\, =\, b,\) and \(\displaystyle y\, +\, z\, =\, c\) have infinitely-many solutions.

from this question i got up to reduced row echelon form of the linear system

that is [1 0 1 a-c; 0 1 1 c; 0 0 0 b-a+c]

but from here I can't figure out the way to solve

are you meant to get exact numbers to a b c

or just specify their ranges

help me on this

 

[Ishuda]

4. Find the values of \(\displaystyle a, \, b,\, c\, \in\, \mathbb{R}\) for which the equations \(\displaystyle x\, +\, y\, +\, 2z\, =\, a,\) \(\displaystyle x\, +\, z\, =\, b,\) and \(\displaystyle y\, +\, z\, =\, c\) have infinitely-many solutions.

from this question i got up to reduced row echelon form of the linear system

that is [1 0 1 a-c; 0 1 1 c; 0 0 0 b-a+c]

but from here I can't figure out the way to solve

are you meant to get exact numbers to a b c

or just specify their ranges

help me on this

Look at that third row. It is equivalent to
0 * x + 0 * y + 0 * z = b - a + c
So, for what values of (b-a+c) is this a valid equation?

 

[sozener1]

Look at that third row. It is equivalent to
0 * x + 0 * y + 0 * z = b - a + c
So, for what values of (b-a+c) is this a valid equation?

b-a+c must be zero in order for this to have infinitely many solutions

but what about other rows???

how do you work out b-a+c using other rows??

 

[Ishuda]

b-a+c must be zero in order for this to have infinitely many solutions

but what about other rows???

how do you work out b-a+c using other rows??

Using the original equations we have
z = b - x
and
z = c - y = b - x
or
z = b - x
y = x + c - b

So, choose x, b and c, let
a = b + c
Then
z = b - x
and
y = x + c - b

 

[sozener1]

Look at that third row. It is equivalent to
0 * x + 0 * y + 0 * z = b - a + c
So, for what values of (b-a+c) is this a valid equation?

b-a+c must be zero in order for this to have infinitely many solutions

but what about other rows???

how do you work out b-a+c using other rows??

Using the original equations we have
z = b - x
and
z = c - y = b - x
or
z = b - x
y = x + c - b

So, choose x, b and c, let
a = b + c
Then
z = b - x
and
y = x + c - b

sorry could you be more clear on what you have provided
becos you seem to have worked out z y and a

but the question asks you to find a b c

do they need to be represented in an algebraic form rather than exact values?? and why??

 

[Ishuda]

sorry could you be more clear on what you have provided
becos you seem to have worked out z y and a

but the question asks you to find a b c

do they need to be represented in an algebraic form rather than exact values?? and why??

Maybe you should read the post again starting at the 'Choose x' or maybe you would prefer what has already been said 'b+c-a=0'

 

[sozener1]

sorry could you be more clear on what you have provided
becos you seem to have worked out z y and a

but the question asks you to find a b c

do they need to be represented in an algebraic form rather than exact values?? and why??

Maybe you should read the post again starting at the 'Choose x' or maybe you would prefer what has already been said 'b+c-a=0'

I get why you get z=b-x and all that
what I don't get is the reason for choosing x b c
and I've been asking you the style of answer for this question should be in??

 

[Steven G]

b-a+c must be zero in order for this to have infinitely many solutions

but what about other rows???

how do you work out b-a+c using other rows??

So if you have infinitely many solutions then can you find out what a, b and c equals??

 

[Steven G]

View attachment 4836

from this question i got up to reduced row echelon form of the linear system

that is [1 0 1 a-c; 0 1 1 c; 0 0 0 b-a+c]

but from here I can't figure out the way to solve

are you meant to get exact numbers to a b c

or just specify their ranges

help me on this

If you have 3 variables with only 2 non zero rows then you will have 3-2 =1 free variables. Let z=r, r is any real number.
The 1st equation says that x+z =a-c or x=-r+a-c
The 2nd equation says that y+z =c or y=-r+c
we know that b-a+c=0 or b=a-c so we can say that x=-r+b

so x=-r+b
y=-r+c
and z=r

Pick a value for r, find x and y, (z=r) and see if this works for each of the 3 equations. Then pick another value for z. Be sure that you see that it works. Even use r for z and it will work!

 

[Steven G]

OK so you want to know what a, b and c can be. That is much much easier. You know that b-a+c=0 or b=a-c. So a and c can be any numbers you like and b must be a-c.
OR
b-a+c=0, so a=b+c. Then b and c can be anything you want and a must equal b+c
OR
c=a-b. Then a and b can be any numbers but c must be a-b.

That is the answer to your question is b-a+c=0 or any thing equivalent to that!