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Date May 2015 Marks available 6 Reference code 15M.2.hl.TZ2.7
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 7 Adapted from N/A

Question

Consider the following system of equations

\[2x + y + 6z = 0\]

\[4x + 3y + 14z = 4\]

\[2x - 2y + (\alpha  - 2)z = \beta  - 12.\]

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.

[6]
a.

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

[3]
b.

Markscheme

\(2x + y + 6z = 0\)

\(4x + 3y + 14z = 4\)

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

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]

a.

\(y + 2z = 4 \Rightarrow y = 4 - 2z\)

\(2x =  - y - 6z = 2z - 4 - 6z =  - 4z - 4 \Rightarrow x =  - 2z - 2\)     A1

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

[3 marks]

Total [9 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1 - Core: Algebra » 1.9 » Solutions of systems of linear equations (a maximum of three equations in three unknowns), including cases where there is a unique solution, an infinity of solutions or no solution.

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