| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix inverse calculation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard techniques: finding when det(M)≠0 for part (a) requires computing a 3×3 determinant, and part (b) uses the standard cofactor/adjugate method for matrix inversion. Part (ii) is a routine proof by induction with simple 2×2 matrix multiplication. All steps are algorithmic with no novel insight required, making it slightly easier than average even for Further Maths content. |
| Spec | 4.01a Mathematical induction: construct proofs4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \( | \mathbf{M} | = 2(1+2)-a(-1-1)+4(2-1)=0 \Rightarrow a = \ldots\) |
| Matrix M has an inverse when \(a \neq -5\) | A1 | Correct condition for \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Minors: \(\begin{pmatrix}3&-2&1\\-a-8&2&a+4\\4-a&-6&-2-a\end{pmatrix}\) or Cofactors: \(\begin{pmatrix}3&2&1\\a+8&2&-a-4\\4-a&6&-2-a\end{pmatrix}\) | B1 | A correct matrix of minors or cofactors |
| \(\mathbf{M}^{-1} = \frac{1}{ | \mathbf{M} | }\text{adj}(\mathbf{M})\) |
| \(\mathbf{M}^{-1} = \frac{1}{2a+10}\begin{pmatrix}3&a+8&4-a\\2&2&6\\1&-a-4&-2-a\end{pmatrix}\) (2 correct rows) | A1ft | Two correct rows following through their detM |
| All correct, following through their detM | A1ft | Fully correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| When \(n=1\), lhs \(= \begin{pmatrix}3&0\\6&1\end{pmatrix}\), rhs \(= \begin{pmatrix}3^1&0\\3(3^1-1)&1\end{pmatrix}=\begin{pmatrix}3&0\\6&1\end{pmatrix}\); true for \(n=1\) | B1 | Shows statement true for \(n=1\) |
| Assume true for \(n=k\): \(\begin{pmatrix}3&0\\6&1\end{pmatrix}^k = \begin{pmatrix}3^k&0\\3(3^k-1)&1\end{pmatrix}\) | M1 | Assumes true for \(n=k\) |
| \(\begin{pmatrix}3&0\\6&1\end{pmatrix}^{k+1} = \begin{pmatrix}3^k&0\\3(3^k-1)&1\end{pmatrix}\begin{pmatrix}3&0\\6&1\end{pmatrix}\) | M1 | Attempts to multiply correct matrices |
| \(= \begin{pmatrix}3\times3^k&0\\3\times3(3^k-1)+6&1\end{pmatrix}\) | A1 | Correct matrix in terms of \(k\) |
| \(= \begin{pmatrix}3^{k+1}&0\\3(3^{k+1}-1)&1\end{pmatrix}\) | A1 | Correct matrix in terms of \(k+1\) |
| If true for \(n=k\) then shown true for \(n=k+1\); true for \(n=1\); therefore true for all positive integers \(n\) | A1 | Correct complete conclusion |
## Question 3(i)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $|\mathbf{M}| = 2(1+2)-a(-1-1)+4(2-1)=0 \Rightarrow a = \ldots$ | M1 | Attempts determinant, equates to zero, solves for $a$ |
| Matrix **M** has an inverse when $a \neq -5$ | A1 | Correct condition for $a$ |
## Question 3(i)(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Minors: $\begin{pmatrix}3&-2&1\\-a-8&2&a+4\\4-a&-6&-2-a\end{pmatrix}$ or Cofactors: $\begin{pmatrix}3&2&1\\a+8&2&-a-4\\4-a&6&-2-a\end{pmatrix}$ | B1 | A correct matrix of minors or cofactors |
| $\mathbf{M}^{-1} = \frac{1}{|\mathbf{M}|}\text{adj}(\mathbf{M})$ | M1 | Complete method for the inverse |
| $\mathbf{M}^{-1} = \frac{1}{2a+10}\begin{pmatrix}3&a+8&4-a\\2&2&6\\1&-a-4&-2-a\end{pmatrix}$ (2 correct rows) | A1ft | Two correct rows following through their det**M** |
| All correct, following through their det**M** | A1ft | Fully correct inverse |
## Question 3(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| When $n=1$, lhs $= \begin{pmatrix}3&0\\6&1\end{pmatrix}$, rhs $= \begin{pmatrix}3^1&0\\3(3^1-1)&1\end{pmatrix}=\begin{pmatrix}3&0\\6&1\end{pmatrix}$; true for $n=1$ | B1 | Shows statement true for $n=1$ |
| Assume true for $n=k$: $\begin{pmatrix}3&0\\6&1\end{pmatrix}^k = \begin{pmatrix}3^k&0\\3(3^k-1)&1\end{pmatrix}$ | M1 | Assumes true for $n=k$ |
| $\begin{pmatrix}3&0\\6&1\end{pmatrix}^{k+1} = \begin{pmatrix}3^k&0\\3(3^k-1)&1\end{pmatrix}\begin{pmatrix}3&0\\6&1\end{pmatrix}$ | M1 | Attempts to multiply correct matrices |
| $= \begin{pmatrix}3\times3^k&0\\3\times3(3^k-1)+6&1\end{pmatrix}$ | A1 | Correct matrix in terms of $k$ |
| $= \begin{pmatrix}3^{k+1}&0\\3(3^{k+1}-1)&1\end{pmatrix}$ | A1 | Correct matrix in terms of $k+1$ |
| If true for $n=k$ then shown true for $n=k+1$; true for $n=1$; therefore true for all positive integers $n$ | A1 | Correct complete conclusion |
---\begin{enumerate}
\item (i)
\end{enumerate}
$$\mathbf { M } = \left( \begin{array} { c c c }
2 & a & 4 \\
1 & - 1 & - 1 \\
- 1 & 2 & - 1
\end{array} \right)$$
where $a$ is a constant.\\
(a) For which values of $a$ does the matrix $\mathbf { M }$ have an inverse?
Given that $\mathbf { M }$ is non-singular,\\
(b) find $\mathbf { M } ^ { - 1 }$ in terms of $a$\\
(ii) Prove by induction that for all positive integers $n$,
$$\left( \begin{array} { l l }
3 & 0 \\
6 & 1
\end{array} \right) ^ { n } = \left( \begin{array} { c c }
3 ^ { n } & 0 \\
3 \left( 3 ^ { n } - 1 \right) & 1
\end{array} \right)$$
\hfill \mbox{\textit{Edexcel CP2 Q3 [12]}}