# Math Help - Solving a set of linear equations problem

1. ## Solving a set of linear equations problem

I have tried and tried to solve this set of 3 linear equations. The problem is that instead of there being 3 equations and 3 unknowns, there are 3 equations with 4 unknowns. I have tried row reduction and keep getting into a mess with squares of the 4th unknown etc. any help would be greatly appreciated.

Q. Solve completely the following set of linear equations.

x + 2y + az = 0
2x + 3y - 2z = a
ax + y + z =3

Cheers,

Shaun.

2. This is more (linear) algebra than calculus, but okey...

Because of the variables used, it looks to me that you have to treat a as a parameter.
First, find out for which values of a, the coëfficiënt matrix isn't regular (hint: think determinant).

3. ok, i've used the coefficient matrix of all the terms on the left hand side, and found that if a=3 or a=-1, then the matrix has a determinant of 0 and therefore no inverse. but i still can't see where this is going. any more help???

4. I found a=-1 or a = 1/3, perhaps one of us made a little mistake.

Now, you have three cases:
1) a = -1
2) a = 1/3
3) a isn't -1 or 1/3

For 1 and 2: substitute the value for a and solve, it'll either be underdetermined or false.
For 3: the system is regular, thus you can uniquely solve x,y,z as a function of a (Cramer, Gauss, ...)

5. Originally Posted by TD!
I found a=-1 or a = 1/3, perhaps one of us made a little mistake.

Now, you have three cases:
1) a = -1
2) a = 1/3
3) a isn't -1 or 1/3

For 1 and 2: substitute the value for a and solve, it'll either be underdetermined or false.
For 3: the system is regular, thus you can uniquely solve x,y,z as a function of a (Cramer, Gauss, ...)
Just for the record, TD's values for a are correct.

-Dan

6. right i don't know how i've managed to get one of the a values correct and not the other. can anyone just go through this question step by step for me, so i can understand the method to reach a solution?

7. I'm sure some here are ready to post a full answer, but I think you'd learn more by trying and asking specific help. Where are you stuck?

8. ok. i made a coefficient matrix with the values of

1 2 a
2 3 -2
a 1 1

and solving i found a to be 3 or -1. so i am confused there.

also having found the values, using Guass does not get me anywhere, i get messed up with values of a^2 etc.

9. Originally Posted by Shaun Gill
right i don't know how i've managed to get one of the a values correct and not the other. can anyone just go through this question step by step for me, so i can understand the method to reach a solution?
$\left | \begin{array}{ccc} 1 & 2 & a \\ 2 & 3 & -2 \\ a & 1 & 1 \end{array} \right | = 1(3*1 - -2*1) - 2(2*1- -2*a) + a(2*1-3*a)$

= $5 - 4 - 4a + 2a - 3a^2 = 0$

$3a^2 + 2a - 1 = (3a - 1)(a + 1) = 0$

Thus a = -1 and a = 1/3 make the determinant 0.

-Dan

10. No, that determinant is 0 for a = 1/3 or a = -1. Using determinant properties:

$
\left| {\begin{array}{*{20}c}
1 & 2 & a \\
2 & 3 & { - 2} \\
a & 1 & 1 \\
\end{array}} \right| = \left| {\begin{array}{*{20}c}
{1 - a} & {2 - a} & a \\
{2 + 2a} & 5 & { - 2} \\
0 & 0 & 1 \\
\end{array}} \right| = \left| {\begin{array}{*{20}c}
{1 - a^2} & {2 - a} \\
{2 + 2a} & 5 \\
\end{array}} \right|
$

So this is: 5(1-a²)-2(1+a)(2-a) = 5-5a²-4-2a+2a² = -3a²-2a+1. This is 0 for a = 1/3 or a = -1.

11. Originally Posted by Shaun Gill
ok. i made a coefficient matrix with the values of

1 2 a
2 3 -2
a 1 1

and solving i found a to be 3 or -1. so i am confused there.

also having found the values, using Guass does not get me anywhere, i get messed up with values of a^2 etc.
The Gaussian method is technically no more difficult here than it is when you have numerical (that is to say non-variable) coefficients.

It IS going to be long and messy. My suggestion is to write out EVERYTHING: every tiny little step by step change that you make. It will take a while, and you'll probably make some mistakes and have to start over, but it's the only way. If you write out all your steps you will be able to find mistakes more easily.

-Dan

12. ok. i'll use the Guassian method then, which will give me expressions for x, y and z in terms of a i suppose? and will the case be that these equations hold for any value of a other than 1/3 and -1???

13. Originally Posted by Shaun Gill
ok. i'll use the Guassian method then, which will give me expressions for x, y and z in terms of a i suppose? and will the case be that these equations hold for any value of a other than 1/3 and -1???
Indeed, and for a = 1/3 and a = -1, you have to check it seperately (as I said: it can either be undetermined or false). Instead of Gauss, Cramer works fine here too.

14. Originally Posted by Shaun Gill
I have tried and tried to solve this set of 3 linear equations. The problem is that instead of there being 3 equations and 3 unknowns, there are 3 equations with 4 unknowns. I have tried row reduction and keep getting into a mess with squares of the 4th unknown etc. any help would be greatly appreciated.

Q. Solve completely the following set of linear equations.

x + 2y + az = 0
2x + 3y - 2z = a
ax + y + z =3

Cheers,

Shaun.

Multiply the first equation by 2 and subtarct the second to get:

(2x+4y+2az)-(2x+3y-2z)=0-a

or:

y + (2a+2)z=-a. ...(A)

Now multiply the second equation by a and the third by 2 and subtarct:

(2ax + 3ay - 2az)-(2ax + 2y + 2z) = a^2-6

or:

(3a-2)y - (2a+2)z = a^2-6 ...(B)

Now add equationa (A) and (B) to get:

(3a-1)y = a^2-a-6,

so:

y = (a^2-a-6)/(3a-1) ...(C).

Substituting this into equation (A) gives:

z=-(4a^2-2a-6)/[(3a-1)(2a+2)]=-[(a+1)(2a-3)]/[(3a-1)(a+1)]

...........................=-(2a-3)/(3a-1) ...(D)

Then substitute (C) and (D) into the first equation to find x.

RonL

(Check the algebra it can always go wrong)

15. I found det(A) before to be -3a²-2a+1. Now, Cramer for x as an example:

$
x = \frac{{\left| {\begin{array}{*{20}c}
0 & 2 & a \\
a & 3 & { - 2} \\
3 & 1 & 1 \\
\end{array}} \right|}}{{\det A}} = \frac{{\left| {\begin{array}{*{20}c}
{ - 3a} & {2 - a} & a \\
{a + 6} & 5 & { - 2} \\
0 & 0 & 1 \\
\end{array}} \right|}}{{\det A}} = \frac{{\left| {\begin{array}{*{20}c}
{ - 3a} & {2 - a} \\
{a + 6} & 5 \\
\end{array}} \right|}}{{\det A}}
$

So:

$
x = \frac{{ - 15a - \left( {2 - a} \right)\left( {a + 6} \right)}}{{ - 3a^2 - 2a + 1}} = \frac{{a^2 - 11a - 12}}{{ - 3a^2 - 2a + 1}} = \frac{{\left( {a + 1} \right)\left( {a - 12} \right)}}{{\left( {a + 1} \right)\left( {1 - 3a} \right)}} = \frac{{a - 12}}{{1 - 3a}}
$

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