5 A matrix \(\mathbf { A }\) has eigenvalues \(- 1,1\) and 2 , with corresponding eigenvectors
$$\left( \begin{array} { r }
0 \\
1 \\
- 2
\end{array} \right) , \quad \left( \begin{array} { r }
- 1 \\
- 1 \\
3
\end{array} \right) \quad \text { and } \quad \left( \begin{array} { r }
2 \\
- 3 \\
5
\end{array} \right) ,$$
respectively. Find \(\mathbf { A }\).
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Question 5:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\mathbf{P} = \begin{pmatrix}0&-1&2\\1&-1&-3\\-2&3&5\end{pmatrix}\), \(\mathbf{D} = \begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}\) B1B1
Identifies matrices P and D
\(\det\mathbf{P} = 1\) B1
Finds inverse of P
\(\mathbf{P}^{-1} = \text{Adj}\,\mathbf{P} = \begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}\) M1A1
\(\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D} \Rightarrow \mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) M1
Uses appropriate result to obtain A
\(\mathbf{A} = \begin{pmatrix}0&-1&2\\-1&-1&-3\\-2&3&5\end{pmatrix}\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}\begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}\) First mark can be implied by correct working
\(= \begin{pmatrix}0&-1&4\\-1&-1&-6\\2&3&10\end{pmatrix}\begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}\) M1A1\(\sqrt{}\)
\(= \begin{pmatrix}3&4&2\\-11&-27&-13\\21&54&26\end{pmatrix}\) A1
Alternative Approach:
Answer Marks
Guidance
Answer/Working Marks
Guidance
Use of \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) (M1)
Obtains 3 sets of 3 linear equations: one set (M1A1)
Other two sets (A1A1)
Solves one set (M1A1)
Solves other sets (A1A1)
(9)
Total: [9]
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## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P} = \begin{pmatrix}0&-1&2\\1&-1&-3\\-2&3&5\end{pmatrix}$, $\mathbf{D} = \begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}$ | B1B1 | Identifies matrices P and D |
| $\det\mathbf{P} = 1$ | B1 | Finds inverse of P |
| $\mathbf{P}^{-1} = \text{Adj}\,\mathbf{P} = \begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}$ | M1A1 | |
| $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D} \Rightarrow \mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$ | M1 | Uses appropriate result to obtain A |
| $\mathbf{A} = \begin{pmatrix}0&-1&2\\-1&-1&-3\\-2&3&5\end{pmatrix}\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}\begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}$ | | First mark can be implied by correct working |
| $= \begin{pmatrix}0&-1&4\\-1&-1&-6\\2&3&10\end{pmatrix}\begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}$ | M1A1$\sqrt{}$ | |
| $= \begin{pmatrix}3&4&2\\-11&-27&-13\\21&54&26\end{pmatrix}$ | A1 | |
**Alternative Approach:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $\mathbf{A}\mathbf{e} = \lambda\mathbf{e}$ | (M1) | |
| Obtains 3 sets of 3 linear equations: one set | (M1A1) | |
| Other two sets | (A1A1) | |
| Solves one set | (M1A1) | |
| Solves other sets | (A1A1) | (9) |
**Total: [9]**
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5 A matrix $\mathbf { A }$ has eigenvalues $- 1,1$ and 2 , with corresponding eigenvectors
$$\left( \begin{array} { r }
0 \\
1 \\
- 2
\end{array} \right) , \quad \left( \begin{array} { r }
- 1 \\
- 1 \\
3
\end{array} \right) \quad \text { and } \quad \left( \begin{array} { r }
2 \\
- 3 \\
5
\end{array} \right) ,$$
respectively. Find $\mathbf { A }$.
\hfill \mbox{\textit{CAIE FP1 2012 Q5 [9]}}