Solving Linear Equations (Matrices)

[Gregg]

Homework Statement

Prove that the equation

\left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; -3 \\<br /> 2 &amp; 6 &amp; -11 \\<br /> 1 &amp; -2 &amp; 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

<br /> \text{Det}\left[\left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; -3 \\<br /> 2 &amp; 6 &amp; -11 \\<br /> 1 &amp; -2 &amp; 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.

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