# Thread: Show that the system of equation is consistent

1. ## Show that the system of equation is consistent

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

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

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

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

3. ## I am still not getting the anw=swer

Originally Posted by mr fantastic
One way of doing this:

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

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

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

I getting the det (A) = 0

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

4. 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.

5. ## Do u have a links for this

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.....

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

CB

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

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

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

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

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

Substitute 4 into equation 3 .

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

$\displaystyle x=2-y$

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

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

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

Not sure if this is right though .

8. ## I couldnt understand

HI

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

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

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

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

Substitute 4 into equation 3 .

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

$\displaystyle x=2-y$

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

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

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

9. Originally Posted by zorro
How did u get $\displaystyle y+2z=3\Rightarrow z=\frac{3-y}{2}$ ---4
((multiply equation 1 by 2) - equation 3) divide by -11

HI

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

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

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

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

Substitute 4 into equation 3 .

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

$\displaystyle x=2-y$

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

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

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

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

the solutions are the set $\displaystyle \{(x,y,z)|(x,y,z)=t(-1,1,-1/2)+(2,0,3/2), t \in \mathbb{R}\}$
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

11. 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.

12. sorry, but i am still unable to understand this ????????????

13. 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

$\displaystyle 0x +0y + 0z = 0$

therefore showing that there is infinite number of solutions