Show that Equations are Consistent

**alexgeek** · September 15th 2010, 10:33 AM

Stuck on a question in a test paper, my book touches this topic but in a different form.
Show that these equations are consistent and find their general solution:
$ \begin{bmatrix} 2&1&1\\ 1&2&1\\ 2&1&1 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} 2\\3\\2 \end{bmatrix} $

Pretty sure I take the determinant.. of something.
Thanks

**yeKciM** · September 15th 2010, 10:36 AM

Originally Posted by alexgeek

Stuck on a question in a test paper, my book touches this topic but in a different form.
Show that these equations are consistent and find their general solution:
$ \begin{bmatrix} 2&1&1\\ 1&2&1\\ 2&1&1 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} 2\\3\\2 \end{bmatrix} $

Pretty sure I take the determinant.. of something.
Thanks

$\begin{bmatrix} 2&1&1\\ 1&2&1\\ 2&1&1 \end{bmatrix}$

if determinant of that is $D \neq 0$ than sys have unique solution ....

Edit:

1° when $D\neq 0$ and if at least one of $D_x\vee D_y\vee D_z\vee \neq 0$ than solutions are unique and given

$\displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}$

2° if $D= 0$ and if at least one of $D_x\vee D_y\vee D_z\vee \neq 0$ than system of equations have no solutions

3° if $D= 0$ and if $D_x= D_y= D_z= 0$ than there can be :

a) all sub determinants of D are zeros, and coefficients with the unknowns are proportional there are 2 cases :

- all independent members of system are proportional with coefficient with unknowns than there are infinity of solutions
- all independent members of system are not proportional with coefficient with unknowns than there are no solutions

b) if at least one sub determinant of D are different from zero than there are infinity of solution

**alexgeek** · September 15th 2010, 11:09 AM

So in the form $\mathbf{AX}=\mathbf{B}$
D is $|\mathbf{A}|$ ?

What are $D_x, D_y, D_z$ ? And where does $\mathbf{B}$ come in?

Thanks!

**yeKciM** · September 15th 2010, 11:26 AM

when you have system of equations .... like :

$a_{11}\cdot x + a_{12}\cdot y +a_{13 }\cdot z = c_1$
$a_{21}\cdot x + a_{22}\cdot y +a_{23 }\cdot z = c_2$
$a_{31}\cdot x + a_{32}\cdot y +a_{33 }\cdot z = c_3$

(do you see such sys in your example ? )

than :

$D= \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix}$

$D_x= \begin{vmatrix} c_{1}&a_{12}&a_{13}\\ c_{2}&a_{22}&a_{23}\\ c_{3}&a_{32}&a_{33} \end{vmatrix}$

$D_y= \begin{vmatrix} a_{11}&c_{1}&a_{13}\\ a_{21}&c_{2}&a_{23}\\ a_{31}&c_{3}&a_{33} \end{vmatrix}$

$D_z= \begin{vmatrix} a_{11}&a_{12}&c_{1}\\ a_{21}&a_{22}&c_{2}\\ a_{31}&a_{32}&c_{3} \end{vmatrix}$

Edit: corrected determinants (first time wrote "bmatrix" sorry for that)

**alexgeek** · September 15th 2010, 11:38 AM

Ah right. So for this question, I would show that all the determinants are non zero and the general solution by the following?
$\displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}$

Thanks

**yeKciM** · September 15th 2010, 11:39 AM

Originally Posted by alexgeek

Ah right. So for this question, I would show that all the determinants are non zero and the general solution by the following?
$\displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}$

Thanks

your determinant D is zero

$D= \begin{vmatrix} 2& 1 & 1 \\ 1&2&1\\ 2&1&1 \end{vmatrix} = 2\cdot 2 \cdot 1 + 1\cdot 1\cdot 2 +1\cdot 1\cdot 1 - ( 1\cdot 2\cdot 2 + 1\cdot 1\cdot 1 +2\cdot 1\cdot 1 ) = 7 -7 = 0$

**alexgeek** · September 15th 2010, 11:51 AM

So does that show that the equations are consistent?
I know you can show they aren't consistent by contradiction, but not sure how to do the reverse.
Thanks, and sorry ha.

**yeKciM** · September 15th 2010, 11:58 AM

Originally Posted by alexgeek

So does that show that the equations are consistent?
I know you can show they aren't consistent by contradiction, but not sure how to do the reverse.
Thanks, and sorry ha.

solve D_x, D_y and D_z as I wrote it to you up there in post #2

and you'll see

you have there $D= D_x=D_y=D_z=0$

so what would you do now ?

Edit: because $D= D_x=D_y=D_z=0$ and there are sub determinants of D that are different from zero, so we know that for your system of equations there is solution , but there are infinity many of them

**alexgeek** · September 15th 2010, 12:12 PM

Hmm, won't you get a divide by zero solving for any of them since D = 0?

**yeKciM** · September 15th 2010, 12:16 PM

Originally Posted by alexgeek

Hmm, won't you get a divide by zero solving for any of them since D = 0?

yes you would if there is any D_x or D_y or D_z different from zero ... this way Cramer rule don't apply because you have D=D_x=D_y=D_z = 0

look at the post #8

**alexgeek** · September 15th 2010, 12:32 PM

Right think I'm almost with you. So does that mean are they linearly independent? Or that all the planes coincide?

And their general solution, is that just $D= D_x=D_y=D_z=0$ ?

God knows how I'm gunna remember all these geometric interpretations for the exam.

Thanks!

**yeKciM** · September 15th 2010, 12:37 PM

Originally Posted by alexgeek

Right think I'm almost with you. So does that mean are they linearly independent? Or that all the planes coincide?

And their general solution, is that just $D= D_x=D_y=D_z=0$ ?

God knows how I'm gunna remember all these geometric interpretations for the exam.

Thanks!

hm....
what is the original question

**alexgeek** · September 15th 2010, 12:46 PM

9. The matrix A is defined by
$\mathbf{A} = \begin{bmatrix} \lambda+1 & 1 & \lambda \\ 1 & 2 & \lambda\\2 & \lambda & 1 \end{bmatrix}$

a. {done this part just proving $\mathbd{A}$ is singular when $\lambda$ = 1}

b. Now consider the system of equations:

$\mathbf{AX}=\mathbf{B}$

where

$\mathbf{X}=\begin{bmatrix}x\\y\\z \end{bmatrix} \; ; \; \mathbf{B} = \begin{bmatrix}2\\3\\2 \end{bmatrix}$

(i) Given that $\lambda = 1$ , show that these equations are consistent and find their general solution.

Just wrote all that and realised I have the PDF open

**yeKciM** · September 15th 2010, 01:12 PM

okay if you are to solve this using inverse matrix there is no solution because there can't be inverse matrix of the matrix A

$AX=B \Rightarrow A^{-1} AX = A^{-1}B \Rightarrow IX= A^{-1}B \Rightarrow X=A^{-1}B$

you can't do inverse matrix if you have determinant of that matrix to be zero

$2x + y+z = 2$
$x + 2y+z = 3$
$2x + y+z = 2$

well now we can go as I wrote it up there, and you'll see that there is infinity of the solutions ... meaning they are dependent... for any x you have different values for y and z for which equations have solutions ...

really don't know which (or how ) did they learned you how to approach to this type of problems ... but in general, same thing (post #2) or at least similar is with homogeneous and inhomogeneous systems of equations (which is easy to remember if you just try little to understand why and how did someone conclude there things ) than you'll be sure at least is there any solutions and what they are

**alexgeek** · September 15th 2010, 01:36 PM

Right, sure I get it now

I'll have to ask my maths teacher what the exam board looks for in answer to questions like this though, not sure what to show and what not to show.
Thanks for all that help!

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