AS October 2020 Paper 1 Q1

EdexcelCurrent spec6 marks3D Lines & Planes Matrices

1. A system of three equations is defined by

\[\begin{aligned}kx + 3y - z &= 3\\ 3x - y + z &= -k\\ -16x - ky - kz &= k\end{aligned}\]

where \(k\) is a positive constant.

Given that there is no unique solution to all three equations,

(a) show that \(k = 2\) (2)

Using \(k = 2\)

(b) determine whether the three equations are consistent, justifying your answer. (3)

Mark scheme (a)

Scheme	Marks	AO
\[\begin{vmatrix}k & 3 & -1\\ 3 & -1 & 1\\ -16 & -k & -k\end{vmatrix} = k(k + k) - 3(-3k + 16) - 1(-3k - 16)\]	M1	2.1
Solves det = 0 \(\Rightarrow 2k^2 + 12k - 32 = 0\) or \(k^2 + 6k - 16 = 0\) To achieve \(k = 2\) (\(k = -8\) must be rejected)	A1	1.1b
	(2)

Notes

M1: Finds the determinant of the matrix corresponding to the system of equations.

A1: Sets determinant = 0 and solves their 3TQ to achieve \(k = 2\) (\(k = -8\) must be rejected)

Special case

Scheme	Marks	AO
\[\begin{vmatrix}2 & 3 & -1\\ 3 & -1 & 1\\ -16 & -2 & -2\end{vmatrix} = 2(2 + 2) - 3(-3 \times 2 + 16) - 1(-3 \times 2 - 16)\]Shows det = 0, therefore when \(k = 2\) there is no unique solution	M1 A0	2.1 1.1b

M1A0: Uses \(k = 2\) and finds the determinant of the matrix corresponding to the system of equations
Shows that determinant = 0 and concludes that when \(k = 2\) there is no unique solution

Mark scheme (b)

Scheme	Marks	AO
Eliminates \(z\) to achieve two equations in \(x\) and \(y\) e.g.\[\begin{aligned}5x + 2y &= 1\\ -10x - 4y &= -2\\ 20x + 8y &= 4\end{aligned}\]or eliminates \(x\) to achieve two equations in \(y\) and \(z\) e.g.\[\begin{aligned}11y - 5z &= 13\\ 22y - 10z &= 26\\ -22y - 10z &= -26\end{aligned}\]or eliminates \(y\) to achieve two equations in \(x\) and \(z\) e.g.\[\begin{aligned}11x + 2z &= -3\\ 22x + 4z &= -6\\ -44x - 8z &= 12\end{aligned}\]	M1 A1	3.1a 1.1b
Must give a reason: e.g. Two equations are a linear multiple of each other e.g. shows they are the same equation therefore the equations are consistent.	A1	2.4
	(3)

Notes

M1: A complete method eliminating one variable from the equations using two different pairs of equations. Condone if a different value of \(k\) is used

A1: Achieves two equations in the same two variables

A1: Must give a reason, shows that the equations are a linear multiple of each other therefore they are consistent.

Alternative

Scheme	Marks	AO
Eliminates two different variables to form two equations, should be one equation from two of the three sections in the main scheme. e.g \(5x + 2y = 1\) and \(11y - 5z = 13\) rearranges and substitutes in to one of the original equations in three variables. e.g. \(2x + 3\left(\dfrac{1 - 5x}{2}\right) - \left(\dfrac{-3 - 11x}{2}\right) = 3\)	M1	3.1a
Correct equations e.g \(5x + 2y = 1\) and \(11y - 5z = 13\)	A1	1.1b
Shows that the equations are a solution e.g. 3 = 3 therefore consistent	A1	2.4

(Note: the substitution shown uses \(5x + 2y = 1\) and \(11x + 2z = -3\).)

M1: A complete method eliminating one variable from the equations using two different pairs of equations. Substitutes these equations into one of the original equations in three variables.

A1: Achieves two correct equations in two different variables

A1: Shows that the equation works therefore they are consistent.

Scheme	Marks	AO
The three planes form a sheaf.	B1	2.2a
	(1)
	(6 marks)