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Date May 2018 Marks available 2 Reference code 18M.1.hl.TZ0.13
Level HL only Paper 1 Time zone TZ0
Command term State Question number 13 Adapted from N/A

Question

Consider the matrix M = \(\left[ {\begin{array}{*{20}{c}}
2 \\
{ - 1}
\end{array}\,\,\,\begin{array}{*{20}{c}}
{ - 4} \\
{ - 1}
\end{array}} \right]\).

Show that the linear transformation represented by M transforms any point on the line \(y = x\) to a point on the same line.

[2]
a.

Explain what happens to points on the line \(4y + x = 0\) when they are transformed by M.

[3]
b.

State the two eigenvalues of M.

[2]
c.

State two eigenvectors of M which correspond to the two eigenvalues.

[2]
d.

Markscheme

\(\left( {\begin{array}{*{20}{c}}
2 \\
{ - 1}
\end{array}\,\,\,\begin{array}{*{20}{c}}
{ - 4} \\
{ - 1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
k \\
k
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 2k} \\
{ - 2k}
\end{array}} \right)\left( { = - 2\left( {\begin{array}{*{20}{c}}
k \\
k
\end{array}} \right)} \right)\)      M1A1

hence still on the line \(y = x\)     AG

[2 marks]

a.

consider \(\left( {\begin{array}{*{20}{c}}
2 \\
{ - 1}
\end{array}\,\,\,\begin{array}{*{20}{c}}
{ - 4} \\
{ - 1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{4k} \\
{ - k}
\end{array}} \right)\)      M1

\( = \left( {\begin{array}{*{20}{c}}
{12k} \\
{ - 3k}
\end{array}} \right)\left( { = 3\left( {\begin{array}{*{20}{c}}
{4k} \\
{ - k}
\end{array}} \right)} \right)\)      A1

hence the line is invariant      A1

[3 marks]

b.

hence the eigenvalues are −2 and 3      A1A1

[2 marks]

c.

\(\left( {\begin{array}{*{20}{c}}
1 \\
1
\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}
4 \\
{ - 1}
\end{array}} \right)\) or equivalent      A1A1

[2 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 1 - Linear Algebra » 1.9 » Eigenvalues and eigenvectors of \(2 \times 2\) matrices.

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