Show that A⁴ = 12A + 5I.

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 $k$.

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]

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]