| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Parameter values for unique solution |
| Difficulty | Standard +0.3 This is a standard Further Maths linear algebra question covering routine techniques: finding parameter values for unique solutions via determinants, verifying no solution cases, finding eigenvalues, and using Cayley-Hamilton theorem for matrix inverse. All are textbook procedures requiring careful calculation but no novel insight, making it slightly easier than average for Further Maths content. |
| Spec | 4.03j Determinant 3x3: calculation4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 1 & -2 & 2 \\ 2 & a-9 & -10 \\ 3 & -6 & 2a \end{vmatrix} = 0\) | M1A1 | Write and evaluate determinant |
| \(2a(a-9) - 60 + 2(4a+30) - 2(-12 - 3a + 27) = 0\) | ||
| \(2a^2 - 4a - 30 = 0\) | M1 | |
| \(a = 5\) or \(a = -3\) | A1 | |
| Unique solution for \(a \neq 5\), \(a \neq -3\) | ||
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1st and 3rd equations: \(x - 2y - 2z + 7 = 0\) | B1 | |
| \(x - 2y - 2z + \frac{29}{3} = 0\) | M1 | |
| Inconsistent, so no solution | ||
| These two equations represent parallel planes. Other equation represents a non-parallel plane which intersects each of the other two in a line. | B1B1 | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 1-\lambda & 1 & 2 \\ 0 & 2-\lambda & 2 \\ -1 & 1 & 3-\lambda \end{vmatrix} = 0\) | B1 | Equate determinant to zero |
| \((1-\lambda)(2-\lambda)(3-\lambda) = 0\) | M1 | Expand determinant and factorise |
| \(\lambda = 1, 2, 3\) | A1A1 | Award A1A0 for 2 correct solutions, A0A0 for 1 correct solution |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A satisfies its characteristic equation, so \(\mathbf{A}^3 - 6\mathbf{A}^2 + 11\mathbf{A} - 6\mathbf{I} = \mathbf{0}\) | B1 | |
| Multiply through by \(\mathbf{A}^{-1}\) to give \(6\mathbf{A}^{-1} = \mathbf{A}^2 - 6\mathbf{A} + 11\mathbf{I}\) | M1 | For information: \(\mathbf{A}^2 = \begin{pmatrix} -1 & 5 & 10 \\ -2 & 6 & 10 \\ -4 & 4 & 9 \end{pmatrix}\) |
| \(\mathbf{A}^{-1} = \frac{1}{6}\begin{pmatrix} 4 & -1 & -2 \\ -2 & 5 & -2 \\ 2 & -2 & 2 \end{pmatrix}\) | M1A1 | |
| Total: 4 marks |
## Question 8(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 1 & -2 & 2 \\ 2 & a-9 & -10 \\ 3 & -6 & 2a \end{vmatrix} = 0$ | M1A1 | Write and evaluate determinant |
| $2a(a-9) - 60 + 2(4a+30) - 2(-12 - 3a + 27) = 0$ | | |
| $2a^2 - 4a - 30 = 0$ | M1 | |
| $a = 5$ or $a = -3$ | A1 | |
| Unique solution for $a \neq 5$, $a \neq -3$ | | |
| **Total: 4 marks** | | |
## Question 8(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1st and 3rd equations: $x - 2y - 2z + 7 = 0$ | B1 | |
| $x - 2y - 2z + \frac{29}{3} = 0$ | M1 | |
| Inconsistent, so no solution | | |
| These two equations represent parallel planes. Other equation represents a non-parallel plane which intersects each of the other two in a line. | B1B1 | |
| **Total: 3 marks** | | |
## Question 8(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 1-\lambda & 1 & 2 \\ 0 & 2-\lambda & 2 \\ -1 & 1 & 3-\lambda \end{vmatrix} = 0$ | B1 | Equate determinant to zero |
| $(1-\lambda)(2-\lambda)(3-\lambda) = 0$ | M1 | Expand determinant and factorise |
| $\lambda = 1, 2, 3$ | A1A1 | Award A1A0 for 2 correct solutions, A0A0 for 1 correct solution |
| **Total: 4 marks** | | |
## Question 8(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **A** satisfies its characteristic equation, so $\mathbf{A}^3 - 6\mathbf{A}^2 + 11\mathbf{A} - 6\mathbf{I} = \mathbf{0}$ | B1 | |
| Multiply through by $\mathbf{A}^{-1}$ to give $6\mathbf{A}^{-1} = \mathbf{A}^2 - 6\mathbf{A} + 11\mathbf{I}$ | M1 | For information: $\mathbf{A}^2 = \begin{pmatrix} -1 & 5 & 10 \\ -2 & 6 & 10 \\ -4 & 4 & 9 \end{pmatrix}$ |
| $\mathbf{A}^{-1} = \frac{1}{6}\begin{pmatrix} 4 & -1 & -2 \\ -2 & 5 & -2 \\ 2 & -2 & 2 \end{pmatrix}$ | M1A1 | |
| **Total: 4 marks** | | |8 (a) (i) Find the set of values of $a$ for which the system of equations
$$\begin{array} { r }
x - 2 y - 2 z + 7 = 0 \\
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q8 [15]}}