Linear systems, unknown coefficient

FrasierCrane

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Joined
Jun 18, 2015
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2
Hi folks, thanks for your time.

I have become stuck on what is probably a very simple question relating to linear systems.

I have three linear equations:
3x+y-2z=-7
-az=x+y+6
2x+2y+z=9​
The coefficient for z in equation 2 is -3, therefore if my maths is correct we can rearrange equation 2 to become x+y-3z= -6.

Solving by gauss-Jordan I have come to the conclusion that x = -2, y = 5, z = 3.
But now I have been asked to find all the values of a, for which this system is consistent, giving algebraic reasoning. I am very stuck here, I have tried gauss-Jordan again, this time replacing the -3 for unknown coefficient a. But with no success. If anyone could shed some light that would be most helpful, cheers :)
 
Hi folks, thanks for your time.

I have become stuck on what is probably a very simple question relating to linear systems.

I have three linear equations:
3x+y-2z=-7
-az=x+y+6
2x+2y+z=9​
The coefficient for z in equation 2 is -3, therefore if my maths is correct we can rearrange equation 2 to become x+y-3z= -6.

Solving by gauss-Jordan I have come to the conclusion that x = -2, y = 5, z = 3.
But now I have been asked to find all the values of a, for which this system is consistent, giving algebraic reasoning. I am very stuck here, I have tried gauss-Jordan again, this time replacing the -3 for unknown coefficient a. But with no success. If anyone could shed some light that would be most helpful, cheers :)

If you have:

[A]{x} = {b}

Then the system is "inconsistent" if |[A]| = 0 while {b}\(\displaystyle \ne 0\)
 
I have three linear equations:
3x+y-2z=-7
-az=x+y+6
2x+2y+z=9​

...I have been asked to find all the values of a, for which this system is consistent....
Restate the equations in the usual way:

. . . . .3x + 1y - 2z = -7
. . . . .1x + 1y + az = -6
. . . . .2x + 2x + 1z = 9

This can be restated in terms of a matrix. Have you worked with matrices at all? Or are you working just with the system of equations?

Thank you! ;)
 
Restate the equations in the usual way:

. . . . .3x + 1y - 2z = -7
. . . . .1x + 1y + az = -6
. . . . .2x + 2x + 1z = 9

This can be restated in terms of a matrix. Have you worked with matrices at all? Or are you working just with the system of equations?

Thank you! ;)

Hi stapel, I have tried to do this in the form of a matrix:

3 1 -2 -7
1 1 a -6
2 2 1 9

But to no avail. I always end up with fractions I can't get rid of.
 
Hi folks, thanks for your time.

I have become stuck on what is probably a very simple question relating to linear systems.

I have three linear equations:
3x+y-2z=-7
-az=x+y+6
2x+2y+z=9​
The coefficient for z in equation 2 is -3, therefore if my maths is correct we can rearrange equation 2 to become x+y-3z= -6.

Solving by gauss-Jordan I have come to the conclusion that x = -2, y = 5, z = 3.
But now I have been asked to find all the values of a, for which this system is consistent, giving algebraic reasoning. I am very stuck here, I have tried gauss-Jordan again, this time replacing the -3 for unknown coefficient a. But with no success. If anyone could shed some light that would be most helpful, cheers :)
If the coefficient of z in the 2nd eq is -3, then that equation becomes -3z = x+y+6. This is NOT the same as x+y-3z=-6. Do you see why?

Just use gauss-jordan to reduce the matrix. Now how were you taught to conclude that a system of of is consistent?? Use that technique. Come back here with and show us your work.
 
Hi folks, thanks for your time.

I have become stuck on what is probably a very simple question relating to linear systems.

I have three linear equations:
3x+y-2z=-7
-az=x+y+6
2x+2y+z=9
The coefficient for z in equation 2 is -3, therefore if my maths is correct we can rearrange equation 2 to become x+y-3z= -6.

Solving by gauss-Jordan I have come to the conclusion that x = -2, y = 5, z = 3.
But now I have been asked to find all the values of a, for which this system is consistent, giving algebraic reasoning. I am very stuck here, I have tried gauss-Jordan again, this time replacing the -3 for unknown coefficient a. But with no success. If anyone could shed some light that would be most helpful, cheers :)

Another method is to use equations 1 and 3 and solve for x and y in terms of z: In matrix form it would be

\(\displaystyle
\begin{pmatrix}
3& 1\\
2& 2
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\begin{pmatrix}
2z-7\\
9-z
\end{pmatrix}
\)

Which is a fairly simple system. After doing that, use equation 2
-az=x+y+6
to determine z in terms of a. That will also tell you what a's, if any, are not allowed.
 
Hi stapel, I have tried to do this in the form of a matrix:

3 1 -2 -7
1 1 a -6
2 2 1 9

But to no avail. I always end up with fractions I can't get rid of.
Fractions are not a bad thing. In fact, they're fairly normal for this sort of exercise.

If you've worked with determinants, try applying what you know about Cramer's Rule (link); in particular, note that the determinant in the denominator had better not be equal to zero. Then set up the equation that Cramer's Rule creates, and solve for the value(s) of "a" which cause the denominator to be zero. Then "a" should be anything else.

Otherwise, just do the solution. Remember that you can't have any row with all zeroes except for the last entry, since this would be an inconsistent system.

For a row-operations solution (rather than a determinant-based solution), I'd start with swapping the first and second rows.

. . . . .\(\displaystyle \left[\begin{array}{cccc}1&1&a&-6\\3&1&-2&-7\\2&2&1&9\end{array}\right]\)

Then add -3*R1 to R3 and -2*R1 to R2. Dividing through to get leading 1's, I get:

. . . . .\(\displaystyle \left[\begin{array}{cccc}1&1&a&-6\\0&1&\frac{2\, +\, 3a}{2}&-\frac{11}{2}\\0&0&1&\frac{21}{1\, -\, 2a}\end{array}\right]\)

Then I'd use the leading 1 in R2 to clear out the 1 in R1:

. . . . .\(\displaystyle \left[\begin{array}{cccc}1&0&\frac{-2\, -\, a}{2}&-\frac{1}{2}\\0&1&\frac{2\, +\, 3a}{2}&-\frac{11}{2}\\0&0&1&\frac{21}{1\, -\, 2a}\end{array}\right]\)

Then use the leading 1 in R3 to clear out the entries in R1 and R2. What can you then conclude?

If you get stuck, please reply showing your steps. Thank you! ;)
 
Hi stapel, I have tried to do this in the form of a matrix:

3 1 -2 -7
1 1 a -6
2 2 1 9

But to no avail. I always end up with fractions I can't get rid of.

|1.....1.......a|
|3.....1.....-2|
|2.....2.......1|

= |1.....-2|........-.........|3.....-2|........+.........a|3.....1|
...|2......1|..................|2.......1|.............,,.....|2.....2|

= (1+ 4) - (3+4) + a(6-2)

=5 - 7 + 4a → the system is inconsistent for a = ½
 
|1.....1.......a|
|3.....1.....-2|
|2.....2.......1|

= |1.....-2|........-.........|3.....-2|........+.........a|3.....1|
...|2......1|..................|2.......1|.............,,.....|2.....2|

= (1+ 4) - (3+4) + a(6-2)

=5 - 7 + 4a → the system is inconsistent for a = ½
Note to original poster: The above is what I'd meant by "using Cramer's Rule" with the denominator's determinant to find the solution. ;)
 
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