| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find inverse then solve system |
| Difficulty | Standard +0.3 This is a straightforward application of standard techniques: finding the inverse of a 3×3 matrix with a parameter (routine cofactor method), then using it to solve a system. While it involves more computation than 2×2 matrices, it requires no problem-solving insight—just careful execution of learned procedures. Slightly above average due to the parameter k and potential for arithmetic errors. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | \mathbf{A} | =3(6-3)-1(6-k)-1(3-k)=2k\) |
| Cofactors matrix: \(\begin{pmatrix}3 & k-6 & 3-k\\-9 & 18+k & k-9\\2 & -4 & 2\end{pmatrix}\) or transpose of matrix of minors: \(\begin{pmatrix}3 & 9 & 2\\6-k & 18+k & 4\\3-k & 9-k & 2\end{pmatrix}\) | M1 | At least 6 correct elements of cofactors or transposed minors |
| \(\mathbf{A}^{-1}=\frac{1}{2k}\begin{pmatrix}3 & -9 & 2\\k-6 & 18+k & -4\\3-k & k-9 & 2\end{pmatrix}\) | dM1, A1 | Complete recognisable method including dividing by determinant; allow minor slips if process clearly correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{2k}\begin{pmatrix}3 & -9 & 2\\k-6 & 18+k & -4\\3-k & k-9 & 2\end{pmatrix}\begin{pmatrix}3\\1\\6\end{pmatrix}=\ldots\) | M1 | Must use their \(\mathbf{A}^{-1}\) in terms of \(k\); condone slip in copying column vector |
| Any 2 of: \(x=\frac{6}{k},\ y=\frac{2k-12}{k},\ z=\frac{6-k}{k}\) | A1 | Two correct simplified or unsimplified expressions; determinant cannot be outside vector |
| \(\left(\frac{6}{k},\frac{2k-12}{k},\frac{6-k}{k}\right)\) or \(x=\frac{6}{k},\ y=2-\frac{12}{k},\ z=\frac{6}{k}-1\) | A1 | Correct coordinates in simplest form; must be written as coordinates not column vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves simultaneously to obtain at least one of \(x\), \(y\), \(z\) in terms of \(k\) | M1 | |
| Two correct expressions for \(x\), \(y\) or \(z\) | A1 | |
| Correct coordinates in simplest form, e.g. \(y=2-\frac{12}{k}\) | A1 | e.g. eliminates \(z\): gets \(4x+2y=4\) and \((6-k)x+3y=0\); uses \(12x+6y=12\) and \((12-2k)x+6y=0\) to get \(2kx=12\Rightarrow x=\frac{6}{k}\) |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|\mathbf{A}|=3(6-3)-1(6-k)-1(3-k)=2k$ | B1 | Correct determinant of $2k$ |
| Cofactors matrix: $\begin{pmatrix}3 & k-6 & 3-k\\-9 & 18+k & k-9\\2 & -4 & 2\end{pmatrix}$ **or** transpose of matrix of minors: $\begin{pmatrix}3 & 9 & 2\\6-k & 18+k & 4\\3-k & 9-k & 2\end{pmatrix}$ | M1 | At least 6 correct elements of cofactors or transposed minors |
| $\mathbf{A}^{-1}=\frac{1}{2k}\begin{pmatrix}3 & -9 & 2\\k-6 & 18+k & -4\\3-k & k-9 & 2\end{pmatrix}$ | dM1, A1 | Complete recognisable method including dividing by determinant; allow minor slips if process clearly correct |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{2k}\begin{pmatrix}3 & -9 & 2\\k-6 & 18+k & -4\\3-k & k-9 & 2\end{pmatrix}\begin{pmatrix}3\\1\\6\end{pmatrix}=\ldots$ | M1 | Must use their $\mathbf{A}^{-1}$ in terms of $k$; condone slip in copying column vector |
| Any 2 of: $x=\frac{6}{k},\ y=\frac{2k-12}{k},\ z=\frac{6-k}{k}$ | A1 | Two correct simplified or unsimplified expressions; determinant cannot be outside vector |
| $\left(\frac{6}{k},\frac{2k-12}{k},\frac{6-k}{k}\right)$ **or** $x=\frac{6}{k},\ y=2-\frac{12}{k},\ z=\frac{6}{k}-1$ | A1 | Correct coordinates in simplest form; must be written as coordinates not column vector |
**Alternative (simultaneous equations):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves simultaneously to obtain at least one of $x$, $y$, $z$ in terms of $k$ | M1 | |
| Two correct expressions for $x$, $y$ or $z$ | A1 | |
| Correct coordinates in simplest form, e.g. $y=2-\frac{12}{k}$ | A1 | e.g. eliminates $z$: gets $4x+2y=4$ and $(6-k)x+3y=0$; uses $12x+6y=12$ and $(12-2k)x+6y=0$ to get $2kx=12\Rightarrow x=\frac{6}{k}$ |
---8.
$$\mathbf { A } = \left( \begin{array} { r r r }
3 & 1 & - 1 \\
1 & 1 & 1 \\
k & 3 & 6
\end{array} \right) \quad k \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $k , \mathbf { A } ^ { - 1 }$
\item Determine, in simplest form in terms of $k$, the coordinates of the point where the following planes intersect.
$$\begin{array} { r }
3 x + y - z = 3 \\
x + y + z = 1 \\
k x + 3 y + 6 z = 6
\end{array}$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 2024 Q8 [7]}}