Solving Linear Equations (Matrices)

[Gregg]

Homework Statement

Prove that the equation

$\left(<br /> \begin{array}{ccc}<br /> 1 & 2 & -3 \\<br /> 2 & 6 & -11 \\<br /> 1 & -2 & 7<br /> \end{array}<br /> \right)\left(<br /> \begin{array}{c}<br /> x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> a \\<br /> b \\<br /> c<br /> \end{array}<br /> \right)$

Is only soluble if $c+2b-5a=0$

(b) Hence show the planes


$x+2y-3z=1$

$2x+6y-11z=2$

$x-2y+7z=1$

Intersect in a line.

(c) Find in terms of s, the co-ordinate of the point in whihc this line meets the plane z=s.

The Attempt at a Solution

$\text{Det}\left[\left(<br /> \begin{array}{ccc}<br /> 1 & 2 & -3 \\<br /> 2 & 6 & -11 \\<br /> 1 & -2 & 7<br /> \end{array}<br /> \right)\right]=0$

$x+2y-3z=a$

$2x+6y-11z=b$

$x-2y+7z=c$

$c+2b-5a = x-2y+7z + 2(2x+6y-11z) -5(x+2y-3z)=0$

Not sure whether this is sufficient?

(b)

$2x+4y-6z=2$

$2x+6y-11z=2$

$y=\frac{5}{2}z$

$x=1-2z$

$z=\lambda$

$x=1-2\lambda, y=\frac{5}{2}\lambda$

What is the vector equation of this result if it is correct?

$r = \left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 0<br /> \end{array}<br /> \right) + \lambda\left(<br /> \begin{array}{c}<br /> -2 \\<br /> \frac{5}{2} \\<br /> 1<br /> \end{array}<br /> \right)$

Is that right?

(c) Substitute z=s

$(1-2s,\frac{5s}{2},s)$

 

[statdad]

You could try reducing the augmented matrix to RREF and looking at what must happen to have a solution.

$$\begin{pmatrix}<br /> 1 & 2 & -3 & a\\<br /> 2 & 6 & -11 & b \\<br /> 1 & -2 & 7 & c<br /> \end{pmatrix}$$