OCR MEI FP2 2011 January — Question 3 16 marks

Exam BoardOCR MEI
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJanuary
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind eigenvalues of 3×3 matrix
DifficultyStandard +0.3 This is a standard FP2 eigenvalue/eigenvector question covering routine techniques: computing a characteristic equation (determinant expansion), verifying an eigenvalue, finding a unit eigenvector, and applying Cayley-Hamilton theorem. While Further Maths content is inherently more advanced, these are textbook procedures with clear algorithms and no novel insight required, placing it slightly above average difficulty overall.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation

  1. Show that the characteristic equation of the matrix $$\mathbf{M} = \begin{pmatrix} 1 & -4 & 5 \\ 2 & 3 & -2 \\ -1 & 4 & 1 \end{pmatrix}$$ is \(\lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0\). [4]
  2. Show that \(\lambda = 3\) is an eigenvalue of \(\mathbf{M}\), and determine whether or not \(\mathbf{M}\) has any other real eigenvalues. [4]
  3. Find an eigenvector, \(\mathbf{v}\), of unit length corresponding to \(\lambda = 3\). State the magnitude of the vector \(\mathbf{M}^n\mathbf{v}\), where \(n\) is an integer. [5]
  4. Using the Cayley-Hamilton theorem, obtain an equation for \(\mathbf{M}^{-1}\) in terms of \(\mathbf{M}^2\), \(\mathbf{M}\) and \(\mathbf{I}\). [3]

(i)
\(\det(M - \lambda I) = (1-\lambda)[(3-\lambda)(1-\lambda) + 8]\)
\(+ 4[2(1-\lambda) - 2] + 5[8 + (3-\lambda)]\)
\(= (1-\lambda)(\lambda^2 - 4\lambda + 11) - 8k + 55 - 5\lambda = 0\)
\(\Rightarrow \lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0\)
AnswerMarks
M1Obtaining \(\det(M - \lambda I)\)
A1Any correct form
M1Simplification
A1 (ag)www, but condone omission of \(= 0\)
Total4
(ii)
\(\lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0\)
\(\Rightarrow (\lambda - 3)(\lambda^2 - 2\lambda + 22) = 0\)
\(\lambda^2 - 2\lambda + 22 = 0 \Rightarrow b^2 - 4ac = -84\)
so no other real eigenvalues
AnswerMarks
M1Factorising and obtaining a quadratic. If M0, give B1 for substituting \(\lambda = 3\)
A1Correct quadratic
M1Considering discriminant o.e.
A1Conclusion from correct evidence www
Total4
(iii)
\(\lambda = 3 \Rightarrow \begin{pmatrix} -2 & -4 & 5 \\ 2 & 0 & -2 \\ -1 & 4 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
\(\Rightarrow -2x - 4y + 5z = 0\)
\(2x - 2z = 0\)
\(-x + 4y - 2z = 0\)
\(\Rightarrow x = z = k, y = \frac{3}{4}k\)
\(\Rightarrow\) eigenvector is \(\begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}\)
\(\Rightarrow\) eigenvector with unit length is \(v = \frac{1}{\sqrt{41}}\begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}\)
Magnitude of \(M^T v\) is \(3^n\)
AnswerMarks
M1Two independent equations
M1Obtaining a non-zero eigenvector
A1
B1
B1Must be a magnitude
Total5
(iv)
\(\lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0\)
\(\Rightarrow M^3 - 5M^2 + 28M - 66I = 0\)
\(\Rightarrow M^2 - 5M + 28I - 66M^{-1} = 0\)
\(\Rightarrow M^{-1} = \frac{1}{66}(M^2 - 5M + 28I)\)
AnswerMarks
M1Use of Cayley-Hamilton Theorem
M1Multiplying by \(M^{-1}\) and rearranging
A1Must contain I
Total3
TOTAL FOR QUESTION 3: 16
## (i)

$\det(M - \lambda I) = (1-\lambda)[(3-\lambda)(1-\lambda) + 8]$
$+ 4[2(1-\lambda) - 2] + 5[8 + (3-\lambda)]$

$= (1-\lambda)(\lambda^2 - 4\lambda + 11) - 8k + 55 - 5\lambda = 0$

$\Rightarrow \lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0$

| | M1 | Obtaining $\det(M - \lambda I)$ |
| | A1 | Any correct form |
| | M1 | Simplification |
| | A1 (ag) | www, but condone omission of $= 0$ |
| **Total** | **4** | |

## (ii)

$\lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0$

$\Rightarrow (\lambda - 3)(\lambda^2 - 2\lambda + 22) = 0$

$\lambda^2 - 2\lambda + 22 = 0 \Rightarrow b^2 - 4ac = -84$

so no other real eigenvalues

| | M1 | Factorising and obtaining a quadratic. If M0, give B1 for substituting $\lambda = 3$ |
| | A1 | Correct quadratic |
| | M1 | Considering discriminant o.e. |
| | A1 | Conclusion from correct evidence www |
| **Total** | **4** | |

## (iii)

$\lambda = 3 \Rightarrow \begin{pmatrix} -2 & -4 & 5 \\ 2 & 0 & -2 \\ -1 & 4 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

$\Rightarrow -2x - 4y + 5z = 0$

$2x - 2z = 0$

$-x + 4y - 2z = 0$

$\Rightarrow x = z = k, y = \frac{3}{4}k$

$\Rightarrow$ eigenvector is $\begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}$

$\Rightarrow$ eigenvector with unit length is $v = \frac{1}{\sqrt{41}}\begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}$

Magnitude of $M^T v$ is $3^n$

| | M1 | Two independent equations |
| | M1 | Obtaining a non-zero eigenvector |
| | A1 | |
| | B1 | |
| | B1 | Must be a magnitude |
| **Total** | **5** | |

## (iv)

$\lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0$

$\Rightarrow M^3 - 5M^2 + 28M - 66I = 0$

$\Rightarrow M^2 - 5M + 28I - 66M^{-1} = 0$

$\Rightarrow M^{-1} = \frac{1}{66}(M^2 - 5M + 28I)$

| | M1 | Use of Cayley-Hamilton Theorem |
| | M1 | Multiplying by $M^{-1}$ and rearranging |
| | A1 | Must contain I |
| **Total** | **3** | |

**TOTAL FOR QUESTION 3: 16**

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\begin{enumerate}[label=(\roman*)]
\item Show that the characteristic equation of the matrix
$$\mathbf{M} = \begin{pmatrix} 1 & -4 & 5 \\ 2 & 3 & -2 \\ -1 & 4 & 1 \end{pmatrix}$$
is $\lambda^3 - 5\lambda^2 + 28\lambda - 66 = 0$. [4]
\item Show that $\lambda = 3$ is an eigenvalue of $\mathbf{M}$, and determine whether or not $\mathbf{M}$ has any other real eigenvalues. [4]
\item Find an eigenvector, $\mathbf{v}$, of unit length corresponding to $\lambda = 3$.

State the magnitude of the vector $\mathbf{M}^n\mathbf{v}$, where $n$ is an integer. [5]
\item Using the Cayley-Hamilton theorem, obtain an equation for $\mathbf{M}^{-1}$ in terms of $\mathbf{M}^2$, $\mathbf{M}$ and $\mathbf{I}$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI FP2 2011 Q3 [16]}}