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]