IB DP Further Mathematics – 1.1 Definition of a matrix: the terms element, row, column HL Paper 1

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Question

The matrix A is given by A = \(\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right)\).

(a)     Given that A\(^3\) can be expressed in the form A\(^3 = a\)A\(^2 = b\)A \( + c\)I, determine the values of the constants \(a\), \(b\), \(c\).

(b)     (i)     Hence express A\(^{ – 1}\) in the form A\(^{ – 1} = d\)A\(^2 = e\)A \( + f\)I where \(d,{\text{ }}e,{\text{ }}f \in \mathbb{Q}\).

(ii)     Use this result to determine A\(^{ – 1}\).

Answer/Explanation

Markscheme

(a)     successive powers of A are given by

A\(^2 = \) \(\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right)\)     (M1)A1

A\(^3 = \) \(\left( {\begin{array}{*{20}{c}}{24}&{35}&{25}\\{25}&{36}&{29}\\{35}&{51}&{36}\end{array}} \right)\)     A1

it follows, considering elements in the first rows, that

\(5a + b + c = 24\)

\(7a + 2b = 35\)

\(6a + b = 25\)     M1A1

solving,     (M1)

\((a,{\text{ }}b,{\text{ }}c) = (3,{\text{ }}7,{\text{ }}2)\)     A1

Note: Accept any other three correct equations.

Note: Accept the use of the Cayley–Hamilton Theorem.

[7 marks]

 

(b)     (i)     it has been shown that

A\(^3 = 3\)A\(^2 + 7\)A\( + 2\)I

multiplying by A\(^{ – 1}\),     M1

A\(^2 = 3\)A\( + 7\)I\( + 2\)A\(^{ – 1}\)     A1

whence

A\(^{ – 1} = 0.5\)A\(^2 – 1.5\)A \( – 3.5\)I     A1

(ii)     substituting powers of A,

A\(^{ – 1} = 0.5\)\(\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right) – 1.5\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right) – 3.5\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\)     M1

=\(\left( {\begin{array}{*{20}{c}}{ – 2.5}&{0.5}&{1.5}\\{1.5}&{ – 0.5}&{ – 0.5}\\{0.5}&{0.5}&{ – 0.5}\end{array}} \right)\)    A1

Note: Follow through their equation in (b)(i).

Note: Line (ii) of (ii) must be seen.

[5 marks]

Question

A matrix M is called idempotent if M\(^2 = \) M.

The idempotent matrix N has the form

N \( = \left( {\begin{array}{*{20}{c}} a&{ – 2a} \\ a&{ – 2a} \end{array}} \right)\)

where \(a \ne 0\).

(i)     Explain why M is a square matrix.

(ii)     Find the set of possible values of det(M).

[4]
a.

(i)     Find the value of \(a\).

(ii)     Find the eigenvalues of N.

(iii)     Find corresponding eigenvectors.

[12]
b.
Answer/Explanation

Markscheme

(i)     M\(^2 = \) MM only exists if the number of columns of M equals the number of rows of M     R1

hence M is square     AG

(ii)     apply the determinant function to both sides     M1

\(\det (\)M\(^2) = \det (\)M\()\)

use the multiplicative property of the determinant

\(\det (\)M\(^2) = \det (\)M\(){\text{ }}\det (\)M\() = \det (\)M\()\)     (M1)

hence \(\det (\)M\() = 0\) or 1     A1

[4 marks]

a.

(i)     attempt to calculate N\(^2\)        M1

obtain \(\left( {\begin{array}{*{20}{c}} { – {a^2}}&{2{a^2}} \\ { – {a^2}}&{2{a^2}} \end{array}} \right)\)        A1

equating to N        M1

to obtain \(a =  – 1\)        A1

(ii)     N \( = \left( {\begin{array}{*{20}{c}} { – 1}&2 \\ { – 1}&2 \end{array}} \right)\)

N \( – \lambda \)I \( = \left( {\begin{array}{*{20}{c}} { – 1 – \lambda }&2 \\ { – 1}&{2 – \lambda } \end{array}} \right)\)        M1

\(( – 1 – \lambda )(2 – \lambda ) + 2 = 0\)       (A1)

\({\lambda ^2} – \lambda  = 0\)       (A1)

\(\lambda \) is 1 or 0        A1

(iii)     let \(\lambda  = 1\)

to obtain \(\left( {\begin{array}{*{20}{c}} { – 1}&2 \\ { – 1}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right){\text{ or }}\left( {\begin{array}{*{20}{c}} { – 2}&2 \\ { – 1}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\)        M1

hence eigenvector is \(\left( {\begin{array}{*{20}{c}} x \\ x \end{array}} \right)\)        A1

let \(\lambda  = 0\)

to obtain \(\left( {\begin{array}{*{20}{c}} { – 1}&2 \\ { – 1}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\)        M1

hence eigenvector is \(\left( {\begin{array}{*{20}{c}} {2y} \\ y \end{array}} \right)\)        A1

Note:     Accept specific eigenvectors.

[12 marks]

b.

Question

Let A2 = 2A + I where A is a 2 × 2 matrix.

Show that A4 = 12A + 5I.

[3]
a.

Let B = \(\left[ {\begin{array}{*{20}{c}}
4&2 \\
1&{ – 3}
\end{array}} \right]\).

Given that B2 – B – 4I = \(\left[ {\begin{array}{*{20}{c}}
k&0 \\
0&k
\end{array}} \right]\), find the value of \(k\).

[3]
b.
Answer/Explanation

Markscheme

METHOD 1
A4 = 4A2 + 4AI + I2 or equivalent M1A1
= 4(2A + I) + 4A + I A1
= 8A + 4I + 4A + I
= 12A + 5I AG

[3 marks]

METHOD 2
A3 = A(2A + I) = 2A2 + AI = 2(2A + I) + A(= 5A + 2I) M1A1
A4 = A(5A + 2I) A1
= 5A2 + 2A = 5(2A + I) + 2A
= 12A + 5I AG

[3 marks]

a.

B2 = \(\left[ {\begin{array}{*{20}{c}}
{18}&2 \\
1&{11}
\end{array}} \right]\) (A1)

\(\left[ {\begin{array}{*{20}{c}}
{18}&2 \\
1&{11}
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
4&2 \\
1&{ – 3}
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
4&0 \\
0&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{10}&0 \\
0&{10}
\end{array}} \right]\) (A1)

\( \Rightarrow k = 10\) A1

[3 marks]

b.

Question

By considering the images of the points (1, 0) and (0, 1),

determine the 2 × 2 matrix P which represents a reflection in the line \(y = \left( {{\text{tan}}\,\theta } \right)x\).

[3]
a.i.

determine the 2 × 2 matrix Q which represents an anticlockwise rotation of θ about the origin.

[2]
a.ii.

Describe the transformation represented by the matrix PQ.

[5]
b.

A matrix M is said to be orthogonal if M TM = I where I is the identity. Show that Q is orthogonal.

[2]
c.
Answer/Explanation

Markscheme

(M1)

using the transformation of the unit square:

\(\left( \begin{gathered}
1 \hfill \\
0 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,2\theta } \\
{{\text{sin}}\,2\theta }
\end{array}} \right)\) and \(\left( \begin{gathered}
0 \hfill \\
1 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{{\text{sin}}\,2\theta } \\
{ – {\text{cos}}\,2\theta }
\end{array}} \right)\) (M1)

hence the matrix P is \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,2\theta } \\
{{\text{sin}}\,2\theta }
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,2\theta } \\
{ – {\text{cos}}\,2\theta }
\end{array}} \right)\) A1

[3 marks]

a.i.

using the transformation of the unit square:

\(\left( \begin{gathered}
1 \hfill \\
0 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}} \right)\) and \(\left( \begin{gathered}
0 \hfill \\
1 \hfill \\
\end{gathered} \right) \to \left( {\begin{array}{*{20}{c}}
{ – {\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\) (M1)

hence the matrix Q is \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{ – {\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\) A1

[2 marks]

a.ii.

PQ = \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta \,{\text{cos}}\,2\theta + {\text{sin}}\,\theta \,{\text{sin}}\,2\theta } \\
{ – {\text{cos}}\,2\theta \,{\text{sin}}\,\theta + \,{\text{sin}}\,2\theta \,{\text{cos}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta \,{\text{sin}}\,2\theta \, – {\text{sin}}\,\theta \,{\text{cos}}\,2\theta } \\
{ – {\text{sin}}\,\theta \,{\text{sin}}\,2\theta – {\text{cos}}\,\theta \,{\text{cos}}\,2\theta \,}
\end{array}} \right)\) M1A1

\( = \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\left( {2\theta – \theta } \right)} \\
{{\text{sin}}\,\left( {2\theta – \theta } \right)}
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,\left( {2\theta – \theta } \right)} \\
{ – {\text{cos}}\,\left( {2\theta – \theta } \right)}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,\theta } \\
{ – {\text{cos}}\,\theta }
\end{array}} \right)\) M1A1

this is a reflection in the line \(y = \left( {{\text{tan}}\,\frac{1}{2}\theta } \right)x\) A1

[5 marks]

b.

Q TQ = \(\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{ – {\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{\text{cos}}\,\theta } \\
{{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{ – {\text{sin}}\,\theta } \\
{{\text{cos}}\,\theta }
\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}
{{\text{co}}{{\text{s}}^2}\,\theta + {\text{si}}{{\text{n}}^2}\,\theta } \\
{ – {\text{sin}}\,\theta \,{\text{cos}}\,\theta + {\text{cos}}\,\theta \,{\text{sin}}\,\theta }
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
{ – {\text{cos}}\,\theta \,{\text{sin}}\,\theta + {\text{cos}}\,\theta \,{\text{sin}}\,\theta } \\
{{\text{si}}{{\text{n}}^2}\,\theta + {\text{co}}{{\text{s}}^2}\,\theta }
\end{array}} \right)\) M1A1

\( = \left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right)\) AG

[2 marks]

c.

Question

The matrix \(\boldsymbol{A}\) is given by \[\boldsymbol{A} = \left( {\begin{array}{*{20}{c}}
0&1&0\\
2&4&1\\
4&{ – 11}&{ – 2}
\end{array}} \right) .\]

(i)     Find the matrices \({\boldsymbol{A}^2}\) and \({\boldsymbol{A}^3}\) , and verify that \({{\boldsymbol{A}}^3} = 2{{\boldsymbol{A}}^2} – {\boldsymbol{A}}\) .

(ii)     Deduce that \({{\boldsymbol{A}}^4} = 3{{\boldsymbol{A}}^2} – 2{\boldsymbol{A}}\) .

[6]
a.

(i)     Suggest a similar expression for \({\boldsymbol{A}^n}\) in terms of \(\boldsymbol{A}\) and \({\boldsymbol{A}^2}\) , valid for \(n \ge 3\) .

(ii)     Use mathematical induction to prove the validity of your suggestion.

[8]
b.
Answer/Explanation

Markscheme

(i)     \({{\boldsymbol{A}}^2} = \left( {\begin{array}{*{20}{c}}
  2&4&1 \\
  4&7&2 \\
  { – 14}&{ – 26}&{ – 7}
\end{array}} \right)\)     A1

\({{\boldsymbol{A}}^3} = \left( {\begin{array}{*{20}{c}}
  4&7&2 \\
  6&{10}&3 \\
  { – 24}&{ – 41}&{ – 12}
\end{array}} \right)\)     A1

\(2{{\boldsymbol{A}}^2} – {\boldsymbol{A}} = 2\left( {\begin{array}{*{20}{c}}
  2&4&1 \\
  4&7&2 \\
  { – 14}&{ – 26}&{ – 7}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  0&1&0 \\
  2&4&1 \\
  { – 4}&{ – 11}&{ – 2}
\end{array}} \right)\)     M1

\( = \left( {\begin{array}{*{20}{c}}
  4&7&2 \\
  6&{10}&3 \\
  { – 24}&{ – 41}&{ – 12}
\end{array}} \right) = {{\boldsymbol{A}}^3}\)     AG

(ii)     \({{\boldsymbol{A}}^4} = {\boldsymbol{A}}{{\boldsymbol{A}}^3}\)     M1

\( = {\boldsymbol{A}}(2{{\boldsymbol{A}}^2} – {\boldsymbol{A}})\)     A1

\( = 2{{\boldsymbol{A}}^3} – {{\boldsymbol{A}}^2}\)

\( = 2(2{{\boldsymbol{A}}^2} – {\boldsymbol{A}}) – {{\boldsymbol{A}}^2}\)     A1

\( = 3{{\boldsymbol{A}}^2} – 2{\boldsymbol{A}}\)     AG

Note: Accept alternative solutions that include correct calculation of both sides of the expression.

[6 marks]

a.

(i)     conjecture: \({{\boldsymbol{A}}^n} = \left( {n – 1} \right){{\boldsymbol{A}}^2} – \left( {n – 2} \right){\boldsymbol{A}}\)     A1

(ii)     first check that the result is true for \(n = 3\)

the formula gives \({{\boldsymbol{A}}^3} = 2{{\boldsymbol{A}}^2} – {\boldsymbol{A}}\) which is correct     A1

assume the result for \(n = k\) , i.e.     M1

\({{\boldsymbol{A}}^k} = (k – 1){{\boldsymbol{{\rm A}}}^2} – (k – 2){\boldsymbol{A}}\)

so

\({{\boldsymbol{A}}^{k + 1}} = {\boldsymbol{A}}\left[ {\left( {k – 1} \right){{\boldsymbol{A}}^2} – \left( {k – 2} \right){\boldsymbol{A}}} \right]\)     M1

\( = \left( {k – 1} \right){{\boldsymbol{A}}^3} – \left( {k – 2} \right){{\boldsymbol{A}}^2}\)     A1

\( = \left( {k – 1} \right)\left( {2{{\boldsymbol{A}}^2} – {\boldsymbol{A}}} \right) – \left( {k – 2} \right){{\boldsymbol{A}}^2}\)     M1

\( = k{{\boldsymbol{A}}^2} – \left( {k – 1} \right){\boldsymbol{A}}\)     A1

so true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 3\) ,

the result is proved by induction     R1

Note: Only award the R1 mark if a reasonable attempt at a proof by induction has been made.

[8 marks]

b.
IB DP Further Mathematics Papers All Topics
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