| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Conditions for unique solution |
| Difficulty | Standard +0.3 This is a straightforward application of determinant conditions for unique solutions. Part (a) requires computing a 3×3 determinant (which will be non-zero for all integer a, making it routine), plus a standard geometric interpretation. Part (b) is simple substitution. The question tests basic understanding of matrix theory without requiring problem-solving insight or complex manipulation. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix4.03t Plane intersection: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} a & 3 & 1 \\ 2 & 1 & 3 \\ -1 & 2 & -5 \end{vmatrix} = -11a + 26\) | M1 A1 | Finds determinant. If solving equations \(x\), \(y\), \(z\) in terms in \(a\). |
| \(\det \mathbf{A} \neq 0\) leads to unique solution. | A1 | States that \(\det \mathbf{A} \neq 0\). (Since \(a\) is an integer.) (M1 A0 A0 if one of \(x\), \(y\), \(z\) found in terms of \(a\).) (M1 A1 A0 if two of \(x\), \(y\), \(z\) found in terms of \(a\).) |
| The three planes intersect at a single point. | B1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 4\) | B1 | |
| Total: 1 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} a & 3 & 1 \\ 2 & 1 & 3 \\ -1 & 2 & -5 \end{vmatrix} = -11a + 26$ | M1 A1 | Finds determinant. If solving equations $x$, $y$, $z$ in terms in $a$. |
| $\det \mathbf{A} \neq 0$ leads to unique solution. | A1 | States that $\det \mathbf{A} \neq 0$. (Since $a$ is an integer.) (M1 A0 A0 if one of $x$, $y$, $z$ found in terms of $a$.) (M1 A1 A0 if two of $x$, $y$, $z$ found in terms of $a$.) |
| The three planes intersect at a single point. | B1 | |
| **Total: 4** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 4$ | B1 | |
| **Total: 1** | | |
---1
\begin{enumerate}[label=(\alph*)]
\item Given that $a$ is an integer, show that the system of equations
$$\begin{aligned}
a x + 3 y + z & = 14 \\
2 x + y + 3 z & = 0 \\
- x + 2 y - 5 z & = 17
\end{aligned}$$
has a unique solution and interpret this situation geometrically.
\item Find the value of $a$ for which $x = 1 , y = 4 , z = - 2$ is the solution to the system of equations in part (a).
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q1 [5]}}