| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A = PDP⁻¹ |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring multiple techniques: proving unique solution via determinant, geometric interpretation, finding eigenvalues from a block-triangular matrix, constructing the diagonalization matrix P from eigenvectors, and using the Cayley-Hamilton theorem to find the inverse. While each individual step is methodical, the combination of parts and the need to work with a parameter 'a' throughout elevates this above a standard question. The diagonalization is straightforward once eigenvalues are found, but part (d) requires knowledge of a less commonly used technique. |
| Spec | 4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} a & 2a+5 & a+1 \\ 0 & -4 & 0 \\ 0 & 3 & -1 \end{vmatrix} = 4a \neq 0\), giving \(x=\frac{11}{2}+\frac{8}{a}\), \(y=-\frac{1}{2}\), \(z=-\frac{9}{2}\) | M1 A1 | Shows determinant is non-zero and solves equations, finds at least two correct values |
| The three planes intersect at a single point | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} a-\lambda & 2a+5 & a+1 \\ 0 & -4-\lambda & 0 \\ 0 & 3 & -1-\lambda \end{vmatrix} = 0\) | M1 | Equates determinant to zero |
| \((a-\lambda)(-4-\lambda)(-1-\lambda)=0 \Rightarrow \lambda = a, -1, -4\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda=a\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -4-a & 0 \\ 0 & 3 & -1-a \end{vmatrix} = \begin{pmatrix}(-4-a)(-1-a)\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda=-1\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a+1&2a+5&a+1\\0&-3&0\end{vmatrix} = \begin{pmatrix}3(a+1)\\0\\-3(a+1)\end{pmatrix} \sim \begin{pmatrix}1\\0\\-1\end{pmatrix}\) | A1 | |
| \(\lambda=-4\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a+4&2a+5&a+1\\0&3&3\end{vmatrix} = \begin{pmatrix}3a+12\\-(3a+12)\\3a+12\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 | |
| \(\mathbf{P} = \begin{pmatrix}1&1&1\\0&0&-1\\0&-1&1\end{pmatrix}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda-a)(\lambda+1)(\lambda+4) = \lambda^3+(5-a)\lambda^2+(4-5a)\lambda-4a=0\) | B1 | Finds characteristic equation |
| \(4a\mathbf{A}^{-1} = \mathbf{A}^2+(5-a)\mathbf{A}+(4-5a)\mathbf{I}\) | M1 A1 | Multiplies through by \(\mathbf{A}^{-1}\) |
| \(\mathbf{A}^2 = \begin{pmatrix}a^2 & 2a^2-17 & a^2-1\\0&16&0\\0&-15&1\end{pmatrix}\) | B1 | |
| \(4a\mathbf{A}^{-1} = \begin{pmatrix}a^2&2a^2-17&a^2-1\\0&16&0\\0&-15&1\end{pmatrix}+(5-a)\begin{pmatrix}a&2a+5&a+1\\0&-4&0\\0&3&-1\end{pmatrix}+(4-5a)\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\) | M1 | Substitutes for \(\mathbf{A}\) and \(\mathbf{A}^2\) in correct equation |
| \(\mathbf{A}^{-1} = \frac{1}{4a}\begin{pmatrix}4&5a+8&4a+4\\0&-a&0\\0&-3a&-4a\end{pmatrix}\) | A1 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} a & 2a+5 & a+1 \\ 0 & -4 & 0 \\ 0 & 3 & -1 \end{vmatrix} = 4a \neq 0$, giving $x=\frac{11}{2}+\frac{8}{a}$, $y=-\frac{1}{2}$, $z=-\frac{9}{2}$ | M1 A1 | Shows determinant is non-zero and solves equations, finds at least two correct values |
| The three planes intersect at a single point | B1 | |
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} a-\lambda & 2a+5 & a+1 \\ 0 & -4-\lambda & 0 \\ 0 & 3 & -1-\lambda \end{vmatrix} = 0$ | M1 | Equates determinant to zero |
| $(a-\lambda)(-4-\lambda)(-1-\lambda)=0 \Rightarrow \lambda = a, -1, -4$ | A1 | AG |
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda=a$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -4-a & 0 \\ 0 & 3 & -1-a \end{vmatrix} = \begin{pmatrix}(-4-a)(-1-a)\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda=-1$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a+1&2a+5&a+1\\0&-3&0\end{vmatrix} = \begin{pmatrix}3(a+1)\\0\\-3(a+1)\end{pmatrix} \sim \begin{pmatrix}1\\0\\-1\end{pmatrix}$ | A1 | |
| $\lambda=-4$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a+4&2a+5&a+1\\0&3&3\end{vmatrix} = \begin{pmatrix}3a+12\\-(3a+12)\\3a+12\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | |
| $\mathbf{P} = \begin{pmatrix}1&1&1\\0&0&-1\\0&-1&1\end{pmatrix}$ | A1 | OE |
## Question 9(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda-a)(\lambda+1)(\lambda+4) = \lambda^3+(5-a)\lambda^2+(4-5a)\lambda-4a=0$ | B1 | Finds characteristic equation |
| $4a\mathbf{A}^{-1} = \mathbf{A}^2+(5-a)\mathbf{A}+(4-5a)\mathbf{I}$ | M1 A1 | Multiplies through by $\mathbf{A}^{-1}$ |
| $\mathbf{A}^2 = \begin{pmatrix}a^2 & 2a^2-17 & a^2-1\\0&16&0\\0&-15&1\end{pmatrix}$ | B1 | |
| $4a\mathbf{A}^{-1} = \begin{pmatrix}a^2&2a^2-17&a^2-1\\0&16&0\\0&-15&1\end{pmatrix}+(5-a)\begin{pmatrix}a&2a+5&a+1\\0&-4&0\\0&3&-1\end{pmatrix}+(4-5a)\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$ | M1 | Substitutes for $\mathbf{A}$ and $\mathbf{A}^2$ in correct equation |
| $\mathbf{A}^{-1} = \frac{1}{4a}\begin{pmatrix}4&5a+8&4a+4\\0&-a&0\\0&-3a&-4a\end{pmatrix}$ | A1 | |9 It is given that $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the system of equations
$$\begin{aligned}
a x + ( 2 a + 5 ) y + ( a + 1 ) z & = 1 \\
- 4 y & = 2 \\
3 y - z & = 3
\end{aligned}$$
has a unique solution and interpret this situation geometrically.\\
The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { c c c }
a & 2 a + 5 & a + 1 \\
0 & - 4 & 0 \\
0 & 3 & - 1
\end{array} \right)$$
\item Show that the eigenvalues of $\mathbf { A }$ are $a , - 1$ and - 4 .
\item Find a matrix $\mathbf { P }$ such that
$$\mathbf { A } = \mathbf { P } \left( \begin{array} { r r r }
a & 0 & 0 \\
0 & - 1 & 0 \\
0 & 0 & - 4
\end{array} \right) \mathbf { P } ^ { - 1 } .$$
\item Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { - 1 }$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q9 [16]}}