Notice $P = I + N$ where $N=\begin{pmatrix}0&0&0\\3&0&0\\9&3&0\end{pmatrix}$ (strictly lower triangular, so $N^3=0$).

$P^5 = (I+N)^5 = I + 5N + 10N^2$ (since $N^3=0$).

$N^2 = \begin{pmatrix}0&0&0\\0&0&0\\9&0&0\end{pmatrix}$... computing: $(N^2)_{31}=3\cdot3=9$, others 0.

$P^5 = I + 5N + 10N^2$. So $Q = P^5 + I = 2I + 5N + 10N^2$.

$q_{21} = 5\cdot3 = 15$, $q_{31} = 5\cdot9+10\cdot9=45+90=135$... wait: $q_{31}=(5N)_{31}+(10N^2)_{31}=5\cdot9+10\cdot9=45+90=135$. Wait $N_{31}=9, N^2_{31}=9$: $q_{31}=5(9)+10(9)=135$. $q_{32}=(5N)_{32}=5\cdot3=15$.

$\dfrac{q_{21}+q_{31}}{q_{32}} = \dfrac{15+135}{15} = \mathbf{10}$

The volume of a cake (same depth) is proportional to the area of the base, which is proportional to the square of the diameter (or radius).

Factor $= \left(\dfrac{12}{8}\right)^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4} = 2\dfrac{1}{4}$

The recipe ingredients should be multiplied by $2\dfrac{1}{4}$.

For identity: $a * e = \frac{ae}{2} = a$, so $ae = 2a$, thus $e = 2$.

• For $He^{+}$, atomic number $Z = 2$.
• The second excited state corresponds to $n = 3$.
• $E_{3} = -13.6 \times \frac{2^{2}}{3^{2}} = -13.6 \times \frac{4}{9}$
• $E_{3} \approx -6.04 \text{ eV}$

The correct behaviour of $g$:

• Inside the Earth ($d < R$): $g = \frac{GM}{R^3}d$ — linear increase from $g=0$ at centre to $g_{surface}$ at $d=R$.
• Outside the Earth ($d > R$): $g = \frac{GM}{d^2}$ — decreases as $1/d^2$.

The graph shows a straight line from origin to $(R, g_{max})$, then a smooth $1/d^2$ curve decreasing afterward. This is a linear rise inside and inverse-square fall outside, meeting at a sharp peak at $d = R$.