A2 June 2025 Paper 2 Q5

Notes

M1: Finds the determinant of the matrix, sets \(= 0\) and solves to find a value of \(p\). Allow for any recognisable attempt at finding the determinant (may be slips in coefficients or signs). Accept if det \(= 14 - 7p\) appears with no working shown.

A1: \(p = 2\) from correct work.

Alt I Using normal

M1: Attempts the cross product between sets of pairs of planes and scales (if appropriate) and equates directions to solve for \(p\).

A1: Correct \(p\)

FYI
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -1 & 1\\ 1 & p & -3\end{vmatrix} = (3 - p)\mathbf{i} + 7\mathbf{j} + (2p + 1)\mathbf{k},\quad \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -1 & 1\\ 3 & 1 & -2\end{vmatrix} = \mathbf{i} + 7\mathbf{j} + 5\mathbf{k},\quad \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & p & -3\\ 3 & 1 & -2\end{vmatrix} = (3 - 2p)\mathbf{i} - 7\mathbf{j} + (1 - 3p)\mathbf{k}\)

Notes

M1: A complete method to find the value of \(q\). May be implied by correct value if no incorrect working is shown.
E.g. Adds together equations 1 and 2 and compares with equation 3 to find a value for \(q\).
Alternatively: Uses equations \(2x - y + z = 3\) and \(3x + y - 2z = 4\) to eliminate one variable. Uses equation \(x + 2y - 3z = q\) and one of the other equations to eliminate the same variable and compare to find a value for \(q\). Another possible method is to identify a point on both planes 1 and 3 (e.g. let \(x = 1\) and solve for \(y\) and \(z\) to get \((1,\ -3,\ -2)\) then substitute into middle equation to find \(q\))

A1: \(q = 1\)

Alternatively II the question may be done as a whole. Eg.
\(\left.\begin{aligned}2x - y + z &= 3\\ 3x + y - 2z &= 4\end{aligned}\right\} \Rightarrow x + 2y - 3z = 1 \Rightarrow p = 2,\ q = 1\) is the most direct method. Variations are possible, e.g.
\(\left.\begin{aligned}2x - y + z &= 3\\ x + py - 3z &= q\\ 3x + y - 2z &= 4\end{aligned}\right\} \Rightarrow \begin{aligned}&3(1) + (2): 7x + (p - 3)y = 9 + q\\ &2(1) + (3): \quad\ 7x - y = 10\end{aligned} \Rightarrow \begin{aligned}&p - 3 = -1 \Rightarrow p = \ldots\\ &9 + q = 10 \Rightarrow q = \ldots\end{aligned}\)

Score as follows:

M1: Attempts to solve at least two of the equations simultaneously to eliminate one variable or match coefficients of the third equation and uses the equation to identify one of the unknowns. Condone slips as long as the method is clear.

A1: Correct \(p\) or correct \(q\).

M1: Full method to use the linear dependence of the equations to find both variables.

A1: Correct \(p\) and \(q\).

Alt III: Using points on the common line.

M1: Finds two separate points on each of the first and third plane.

A1: Two correct points.

M1: Forms and solves two equations in \(p\) and \(q\) using their points.

A1: Correct \(p\) and \(q\).

E.g.
\(x = 0 \Rightarrow \begin{cases}-y + z = 3\\ y - 2z = 4\end{cases} \Rightarrow y = -10,\ z = -7\)
\(x = 1 \Rightarrow \begin{cases}-y + z = 1\\ y - 2z = 1\end{cases} \Rightarrow y = -3,\ z = -2\)
\(\Rightarrow \begin{aligned}-10p + 21 &= q\\ 1 - 3p + 6 &= q\end{aligned} \Rightarrow p = \ldots,\ q = \ldots\)

Some useful equations:

Eliminating \(x\): \(\begin{cases}(2p + 1)y - 7z = 2q - 3\\ 5y - 7z = -1\\ (3p - 1)y - 7z = 3q - 4\end{cases}\)

Eliminating \(y\): \(\begin{cases}(2p + 1)x + (p - 3)z = 3p + q\\ 5x - z = 7\\ (3p - 1)x + (3 - 2p)z = 4p - q\end{cases}\)

Eliminating \(z\): \(\begin{cases}7x + (p - 3)y = 9 + q\\ 7x - y = 10\\ 7x + (3 - 2p)y = 12 - 2q\end{cases}\)