# Solving Linear Equations (Matrices)

1. Jun 16, 2009

### Gregg

1. The problem statement, all variables and given/known data

Prove that the equation

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

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.

3. The attempt at a solution
$\text{Det}\left[\left( \begin{array}{ccc} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{array} \right)\right]=0$

$x+2y-3z=a$

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

$x-2y+7z=c$

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

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?

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

Is that right?

(c) Substitute z=s

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

2. Jun 16, 2009

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