19M.1.AHL.TZ0.F_3

pestleMathematicsAIHLPaper 119M· ahl-1-15-eigenvalues-and-eigenvectorssource ↗

The matrix A is given by A $= \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]$ .

The matrix B is given by B $= \left[ {\begin{array}{*{20}{c}} 3&2 \\ 2&3 \end{array}} \right]$ .

Show that the eigenvalues of A are real if ${\left( {a - d} \right)^2} + 4bc \geqslant 0$ .

[4]

a.i.

Deduce that the eigenvalues are real if A is symmetric.

[2]

a.ii.

Determine the eigenvalues of B.

[2]

b.i.

Determine the corresponding eigenvectors.

[4]

b.ii.

Markscheme / solution

the eigenvalues satisfy

$\left| {\begin{array}{*{20}{c}} {a - \lambda }&b \\ c&{d - \lambda } \end{array}} \right| = 0$ M1

$\left( {a - \lambda } \right)\left( {d - \lambda } \right) - bc = 0$ A1

${\lambda ^2} - \left( {a + d} \right)\lambda + ad - bc = 0$ A1

the condition for real roots is

${\left( {a + d} \right)^2} - 4\left( {ad - bc} \right) \geqslant 0$ M1

${\left( {a - d} \right)^2} + 4bc \geqslant 0$ AG

[4 marks]

a.i.

if the matrix is symmetric, b = c. In this case, M1

${\left( {a - d} \right)^2} + 4bc = {\left( {a - d} \right)^2} + 4{b^2} \geqslant 0$

because each square term is non-negative R1AG

[2 marks]

a.ii.

the characteristic equation is

${\lambda ^2} - 6\lambda + 5 = 0$ M1

$\lambda = 1,5$ A1

[2 marks]

b.i.

taking $\lambda = 1$

$\left[ {\begin{array}{*{20}{c}} 2&2 \\ 2&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right]$ M1

giving eigenvector $= \left[ {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right]$ A1

taking $\lambda = 5$

$\left[ {\begin{array}{*{20}{c}} { - 2}&2 \\ 2&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right]$ M1

giving eigenvector $= \left[ {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right]$ A1

[4 marks]

b.ii.

Examiners’ report

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