# Math Help - Solving a system of linear equations using matrices.

1. ## Solving a system of linear equations using matrices.

the cartesian equations of 3 planes are:
x + y + z = 3
3x - y - z = 1
x + py = q

By writing down the augmented matrix for this system and reducing it to echelon form,
find for what values of p and q these planes
(a) intersect at a single point;
(b) intersect along a line;
(c) have no common line or point of intersection.
In case (a) find, in terms of p and q, the coordinates of the point, and in case (b) find the scalar parametric equations of the line.

i got the row echelon form to be
1 2 2 | 3
0 1 2 | 2
0 0 2p | q+1+2p

is the above true and if so are these true?
case A)
z = (q+1+2p)/2p
y=2-(2q+2+4p)/2p
x=1

case B) DONT UNDERSTAND

case C) at p = 0 ; q diff than -1

Im soooo confused.. ALL HELP APPRECIATED!

2. Originally Posted by swiftshift
the cartesian equations of 3 planes are:
x + y + z = 3
3x - y - z = 1
x + py = q

By writing down the augmented matrix for this system and reducing it to echelon form,
find for what values of p and q these planes
(a) intersect at a single point;
(b) intersect along a line;
(c) have no common line or point of intersection.
In case (a) find, in terms of p and q, the coordinates of the point, and in case (b) find the scalar parametric equations of the line.

i got the row echelon form to be
1 2 2 | 3
0 1 2 | 2
0 0 2p | q+1+2p

is the above true and if so are these true?
case A)
z = (q+1+2p)/2p
y=2-(2q+2+4p)/2p
x=1

case B) DONT UNDERSTAND

case C) at p = 0 ; q diff than -1

Im soooo confused.. ALL HELP APPRECIATED!
$\begin{bmatrix}
1 & 1 & 1 & 3\\
3 & -1 & -1 & 1\\
1 & p & 0 & q
\end{bmatrix}$

$ref=\begin{bmatrix}
1 & 1 & 1 & 3\\
0 & 1 & 1 & 2\\
0 & 0 & p & 2p+1-q
\end{bmatrix}$

Part a is when $p=1$ and $q\in\mathbb{R}$
Part b (still thinking)
Part c is when $p=0$ and $q\in\mathbb{R}, q\neq1$

3. For any system of equations, there are 3 possibilities: there is a unique solution, there is NO solution, or there exist an infinite number of solutions.

Your row reduce matrix is $\begin{bmatrix}
1 & 1 & 1 & 3\\
0 & 1 & 1 & 2\\
0 & 0 & p & 2p+1-q
\end{bmatrix}$

which is equivalent to the equations x+ y+ z= 3, y+ z= 2, and pz= 2p+1- q.

As long as $p\ne 0$, we can divide both sides of the last equation by p to get $z= \frac{2p+ 1- q}{p}$, then use that value in y+ z= 2 to solve for y, then use those in x+ y+ z= 3 to solve for x.

If p= 0 and $2p+ 1- q= 1- q\ne 0$, then the last equation is
0z= 2p+ 1- q which is impossible- 0 times any number is 0. There is no solution, the three planes do not intersect.

If p= 0 and 2p+ 1- q= 1- q= 0[/tex], that is, if p= 0 and q= 1, the last equation is 0z= 0 which is true for all z. In that case, the second equation becomes y+ z= 2 or y= 2- z. The first equation, x+ y+ z= 3 becomes x+ 2= 3 so x= 1. (1, 2- z, z) for any z will satisfy the equations. Any point along that line (in parametric equations, x= 1, y= 2- t, z= t) will satisfy those equations so the three planes intersect on that line.