Find conditions on $\alpha$ and $\beta$ for which

(i)     the system has no solutions;

(ii)     the system has only one solution;

(iii)     the system has an infinite number of solutions.

In the case where the number of solutions is infinite, find the general solution of the system of equations in Cartesian form.

$2x + y + 6z = 0$

$4x + 3y + 14z = 4$

$2x - 2y + (\alpha  - 2)z = \beta  - 12$

attempt at row reduction     M1

eg$\;\;\;{R_2} - 2{R_1}{\text{ and }}{R_3} - {R_1}$

$2x + y + 6z = 0$

$y + 2z = 4$

$- 3y + (\alpha  - 8)z = \beta  - 12$     A1

eg$\;\;\;{R_3} + 3{R_2}$

$2x + y + 6z = 0$

$y + 2z = 4$     A1

$(\alpha  - 2)z = \beta$

(i)     no solutions if $\alpha  = 2,{\text{ }}\beta  \ne 0$     A1

(ii)     one solution if $\alpha  \ne 2$     A1

(iii)     infinite solutions if $\alpha  = 2,{\text{ }}\beta  = 0$     A1

Note:     Accept alternative methods e.g. determinant of a matrix

Note:     Award A1A1A0 if all three consistent with their reduced form, A1A0A0 if two or one answer consistent with their reduced form.

[6 marks]

$y + 2z = 4 \Rightarrow y = 4 - 2z$

$2x =  - y - 6z = 2z - 4 - 6z =  - 4z - 4 \Rightarrow x =  - 2z - 2$     A1

therefore Cartesian equation is $\frac{{x + 2}}{{ - 2}} = \frac{{y - 4}}{{ - 2}} = \frac{z}{1}$ or equivalent     A1

[3 marks]

Total [9 marks]