OCR FP1 2011 January — Question 9 9 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeParameter values for unique solution
DifficultyStandard +0.3 This is a standard Further Maths question testing routine determinant calculation, inverse existence conditions, and applying det≠0 for unique solutions. Part (iii) requires recognizing that a=6 makes det(M)=0, but the calculation is straightforward. Slightly easier than average A-level due to being mechanical application of learned techniques.
Spec4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03s Consistent/inconsistent: systems of equations

9 The matrix \(\mathbf { M }\) is given by \(\mathbf { M } = \left( \begin{array} { r r r } a & - a & 1 \\ 3 & a & 1 \\ 4 & 2 & 1 \end{array} \right)\).
  1. Find, in terms of \(a\), the determinant of \(\mathbf { M }\).
  2. Hence find the values of \(a\) for which \(\mathbf { M } ^ { - 1 }\) does not exist.
  3. Determine whether the simultaneous equations $$\begin{aligned} & 6 x - 6 y + z = 3 k \\ & 3 x + 6 y + z = 0 \\ & 4 x + 2 y + z = k \end{aligned}$$ where \(k\) is a non-zero constant, have a unique solution, no solution or an infinite number of solutions, justifying your answer.
  4. Show that \(\frac { 1 } { r } - \frac { 2 } { r + 1 } + \frac { 1 } { r + 2 } \equiv \frac { 2 } { r ( r + 1 ) ( r + 2 ) }\).
  5. Hence find an expression, in terms of \(n\), for $$\sum _ { r = 1 } ^ { n } \frac { 2 } { r ( r + 1 ) ( r + 2 ) }$$
  6. Show that \(\sum _ { r = n + 1 } ^ { \infty } \frac { 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { ( n + 1 ) ( n + 2 ) }\).

Question 9:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
M1Show correct expansion process for \(3\times3\)
M1Correct evaluation of any \(2\times2\)
\(\det M = a^2 - 7a + 6\)A1 3 Correct answer
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
M1Solve \(\det M = 0\)
\(a = 1\) or \(6\)A1 A1 3 Obtain correct answer, ft their (i)
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
M1Attempt to eliminate one variable
A1Obtain 2 correct equations in 2 unknowns
A1 3Justify infinite number of solutions; SC 3/3 if unique solution conclusion consistent with their (i) or (ii)
## Question 9:

**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Show correct expansion process for $3\times3$ |
| | M1 | Correct evaluation of any $2\times2$ |
| $\det M = a^2 - 7a + 6$ | A1 **3** | Correct answer |

**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Solve $\det M = 0$ |
| $a = 1$ or $6$ | A1 A1 **3** | Obtain correct answer, ft their (i) |

**Part (iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Attempt to eliminate one variable |
| | A1 | Obtain 2 correct equations in 2 unknowns |
| | A1 **3** | Justify infinite number of solutions; SC 3/3 if unique solution conclusion consistent with their (i) or (ii) |

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9 The matrix $\mathbf { M }$ is given by $\mathbf { M } = \left( \begin{array} { r r r } a & - a & 1 \\ 3 & a & 1 \\ 4 & 2 & 1 \end{array} \right)$.\\
(i) Find, in terms of $a$, the determinant of $\mathbf { M }$.\\
(ii) Hence find the values of $a$ for which $\mathbf { M } ^ { - 1 }$ does not exist.\\
(iii) Determine whether the simultaneous equations

$$\begin{aligned}
& 6 x - 6 y + z = 3 k \\
& 3 x + 6 y + z = 0 \\
& 4 x + 2 y + z = k
\end{aligned}$$

where $k$ is a non-zero constant, have a unique solution, no solution or an infinite number of solutions, justifying your answer.\\
(i) Show that $\frac { 1 } { r } - \frac { 2 } { r + 1 } + \frac { 1 } { r + 2 } \equiv \frac { 2 } { r ( r + 1 ) ( r + 2 ) }$.\\
(ii) Hence find an expression, in terms of $n$, for

$$\sum _ { r = 1 } ^ { n } \frac { 2 } { r ( r + 1 ) ( r + 2 ) }$$

(iii) Show that $\sum _ { r = n + 1 } ^ { \infty } \frac { 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { ( n + 1 ) ( n + 2 ) }$.

\hfill \mbox{\textit{OCR FP1 2011 Q9 [9]}}