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Date May 2014 Marks available 6 Reference code 14M.2.hl.TZ1.4
Level HL only Paper 2 Time zone TZ1
Command term Find and Show that Question number 4 Adapted from N/A

Question

A system of equations is given below.

     \(x + 2y - z = 2\)

     \(2x + y + z = 1\)

     \( - x + 4y + az = 4\)

(a)     Find the value of a so that the system does not have a unique solution.

(b)     Show that the system has a solution for any value of a.

Markscheme

(a)   \(\left[ \begin{array}{l}x + 2y - z = 2\\2x + y + z = 1\\ - x + 4y + az = 4\end{array} \right.\)

 

\( \to \left[ \begin{array}{l}x + 2y - z = 2\\ - 3y + 3z =  - 3\\6y + (a - 1)z = 6\end{array} \right.\)     M1A1

 

\( \to \left[ \begin{array}{l}x + 2y - z = 2\\ - 3y + 3z =  - 3\\(a + 5)z = 0\end{array} \right.\)     A1

(or equivalent)

if not a unique solution then \(a =  - 5\)     A1

 

Note:     The first M1 is for attempting to eliminate a variable, the first A1 for obtaining two expression in just two variables (plus a), and the second A1 for obtaining an expression in just a and one other variable

 

[4 marks]

 

(b)     if \(a =  - 5\) there are an infinite number of solutions as last equation always true     R1

and if \(a \ne  - 5\) there is a unique solution     R1

hence always a solution     AG

[2 marks]

 

Total [6 marks]

Examiners report

[N/A]

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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