# Show that the system of equation is consistent

• May 3rd 2009, 10:24 PM
zorro
Show that the system of equation is consistent
Show that the system of equation $
x - 3y -8z = -10;
$
$
3x+y-4z=0;
$
and $
2x+5y+6z =13
$
is consistent and find the solution .
• May 4th 2009, 12:30 AM
mr fantastic
Quote:

Originally Posted by zorro
Show that the system of equation $
x - 3y -8z = -10;
$
$
3x+y-4z=0;
$
and $
2x+5y+6z =13
$
is consistent and find the solution .

One way of doing this:

Express the sytem in matrix form: $A X = B$.

Construct the coefficient matrix A and show that the $\det(A) \neq 0$.

Solve $A X = B$: $A X = B \Rightarrow X = A^{-1} B$.
• Dec 14th 2009, 01:39 PM
zorro
I am still not getting the anw=swer
Quote:

Originally Posted by mr fantastic
One way of doing this:

Express the sytem in matrix form: $A X = B$.

Construct the coefficient matrix A and show that the $\det(A) \neq 0$.

Solve $A X = B$: $A X = B \Rightarrow X = A^{-1} B$.

I getting the det (A) = 0

I dont know what to do??????
• Dec 14th 2009, 05:36 PM
mr fantastic
Quote:

Originally Posted by zorro
I getting the det (A) = 0

I dont know what to do??????

OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

In fact, since x = 3y + 8z from the first equation, the second and third equations become:

y + 2z = 0 and y + 2z = 13/2.

Therefore the equations are INconsistent and there is no solution to be found.
• Dec 19th 2009, 08:36 PM
zorro
Do u have a links for this
Quote:

Originally Posted by mr fantastic
OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

In fact, since x = 3y + 8z from the first equation, the second and third equations become:

y + 2z = 0 and y + 2z = 13/2.

Therefore the equations are INconsistent and there is no solution to be found.

Mr fantastic could u please provide me with a link for solving such type of question's.....
• Dec 19th 2009, 11:02 PM
CaptainBlack
Quote:

Originally Posted by mr fantastic
OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

In fact, since x = 3y + 8z from the first equation, the second and third equations become:

y + 2z = 0 and y + 2z = 13/2.

Therefore the equations are INconsistent and there is no solution to be found.

Unfortunately $x=1$, $y=1$, $z=1$ IS a solution (So the system is consistent as at least one solution exists. The given solution is not unique!).

CB
• Dec 19th 2009, 11:54 PM
Quote:

Originally Posted by zorro
Show that the system of equation $
x - 3y -8z = -10;
$
$
3x+y-4z=0;
$
and $
2x+5y+6z =13
$
is consistent and find the solution .

HI

$x-3y-8z=-10$ ---- 1

$3x+y-4z=0$ -----2

$2x+5y+6z=13$ ---- 3

1x2 -3 $y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4

Substitute 4 into equation 3 .

$2x+5y+6(\frac{3-y}{2})=13$

$x=2-y$

so (x,y,z)=(2-y , y , (3-y)/2)

(x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where $y\in R$

and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .

Not sure if this is right though .
• Dec 20th 2009, 01:02 AM
zorro
I couldnt understand
Quote:

HI

$x-3y-8z=-10$ ---- 1

$3x+y-4z=0$ -----2

$2x+5y+6z=13$ ---- 3

1x2 -3 $y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4

Substitute 4 into equation 3 .

$2x+5y+6(\frac{3-y}{2})=13$

$x=2-y$

so (x,y,z)=(2-y , y , (3-y)/2)

(x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where $y\in R$

and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .

Not sure if this is right though .

How did u get $y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4
• Dec 20th 2009, 01:42 AM
dedust
Quote:

Originally Posted by zorro
How did u get $y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4

((multiply equation 1 by 2) - equation 3) divide by -11

Quote:

HI

$x-3y-8z=-10$ ---- 1

$3x+y-4z=0$ -----2

$2x+5y+6z=13$ ---- 3

1x2 -3 $y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4

Substitute 4 into equation 3 .

$2x+5y+6(\frac{3-y}{2})=13$

$x=2-y$

so (x,y,z)=(2-y , y , (3-y)/2)

(x,y,z)=y(-1,1,-1/2)+(2,0,3/2) where $y\in R$

and if y=1 , (x,y,z)=(1,1,1) which agrees with CB .

Not sure if this is right though .

the solutions are the set $\{(x,y,z)|(x,y,z)=t(-1,1,-1/2)+(2,0,3/2), t \in \mathbb{R}\}$
there's an infinite number of solutions
• Dec 20th 2009, 01:45 AM
CaptainBlack
Quote:

Originally Posted by dedust
((multiply equation 1 by 2) - equation 3) divide by -11

the solutions are the set $\{(x,y,z)|(x,y,z)=t(-1,1,-1/2)+(2,0,3/2), t \in \mathbb{R}\}$
there's an infinite number of solutions

All of which is just of just academic interest here, to show consistency it is sufficient to show that there is at least one solution.

CB
• Dec 20th 2009, 02:58 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
OK, the determinamt is equal to zero (I should have checked). This means that either there's no solution or an infinite number of solutions. So my suggestion is no good after all.

In fact, since x = 3y + 8z -10 from the first equation, Mr F says: An embarrassing correction in red. So what follows is wrong.

the second and third equations become:

y + 2z = 0 and y + 2z = 13/2.

Therefore the equations are INconsistent and there is no solution to be found.

Therefore the second and third equations become y + 2z = 3 and y + 2z = 3. Therefore there are an infinite number of solutions (the one given by CB is once such). These solutions can be expressed in parametric form. The system is consistent.
• Dec 27th 2009, 01:09 AM
zorro
sorry, but i am still unable to understand this ????????????
• Dec 27th 2009, 01:44 AM
11rdc11
I'm not sure if this is a right approach to this problem due to teaching myself the material but could you not just use gauss-jordan elimination.

The last row would be

$0x +0y + 0z = 0$

therefore showing that there is infinite number of solutions