1. ## Matricies

The variables x, y, z satisfy the equations

x + 2y + 3y = 2
x + 3y + 2z = 1
2x + 2y + 𝛌z = 𝛍

where, 𝛌, 𝛍 are constants

a) using reduction to echelon form, or otherwise,

i) Find the value of 𝛌 for which these equations do not have a unique solution
ii) for this value of 𝛌, find the value of 𝛍 for which the equations are consistent.

b) for your values of 𝛌,𝛍, find the general solution of these equations.

I have managed to reduce the equations to

x + 2y + 3z = 2
0 - y + z = 1
0 + 0 + 8-𝛌 = 6-𝛍

I'm not sure what to do next as i'm not sure what it means 'or which these equations do not have a unique solution'

Has anybody got any ideas?

thanks

2. Hello, djmccabie!

Couldn't you use letters for the constants?

$\begin{array}{ccc}x + 2y + 3y &=& 2 \\x + 3y + 2z &=& 1 \\ 2x + 2y + az &=& b \end{array}$

a) Using reduction to echelon form, or otherwise,

i) Find the value of $a$ for which these equations do not have a unique solution.

We have: . $\left[ \begin{array}{ccc|c}1&2&3&2\\1&3&2&1\\2&2&a&b \end{array} \right]\qquad{\color{blue}(1)}$

$\begin{array}{c}\\R_2-R_1 \\ R_3-2R_1\end{array} \left[\begin{array}{ccc|c}1 &2&3&2 \\ 0&1&\text{-}1&\text{-}1 \\ 0 & \text{-}2 &a-6 & b-4 \end{array}\right] \qquad{\color{blue}(2)}$

$\begin{array}{c}R_1-2R_2 \\ \\ R_3+2R_2\end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&1& \text{-}1 & \text{-}1 \\ 0 & 0 & a-4 & b-6 \end{array} \right] \qquad{\color{blue}(3)}$

. . $\begin{array}{c}\\ \\ \frac{1}{a-4}R_3 \end{array} \left[\begin{array}{ccc|c}1 & 0 & 5 & 4 \\ 0 & 1 & \text{-}1 & \text{-}1 \\ 0 & 0 & 1 & \frac{b-6}{a-4}\end{array} \right] \qquad{\color{blue}(4)}$

$\begin{array}{c}R_1-5R_3 \\ R_2 + R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1 &0&0 & \frac{4a-5b-14}{a-4} \\ \\[-4mm] 0&1&0 & \frac{b-a-2}{a-4} \\ \\[-4mm] 0&0&1&\frac{b-6}{a-4} \end{array}\right] \qquad{\color{blue}(5)}$

If $a = 4$, the system does not have a unique solution.

ii) For this value of $a$, find the value of $b$ for which the equations are consistent.

If $a = 4$, then $b = 6$ produces a system with an infinite number of solutions.

(If $b \neq 6$, the system has no solution.)

b) For your values of $a,b$, find the general solution of these equations.

If $a = 4$ and $b = 6$, then [3] becomes: . $\left[ \begin{array}{ccc|c}1&0&5&4 \\ 0&1&\text{-}1 & \text{-}1 \\ 0&0&0&0 \end{array}\right]$

And we have: . $\begin{array}{c}x+5z \:=\:4 \\ y - z \:=\:\text{-}1 \end{array}$

. . which can be written: . $\begin{array}{ccc}x &=&4 - 5z \\ y &=& z - 1 \\ z &=& z\end{array}$

On the right, replace $z$ with a parameter $t\!:\quad \begin{array}{ccc}x &=& 4 - 5t \\ y &=& t - 1 \\ z &=&t \end{array}$

This represents all the solutions to this system of equations.