Matrix (JEE)

$P= \begin{bmatrix}1 & 0& 0\\ 4& 1 & 0\\ 16 & 4 & 1 \end{bmatrix},$ $P^2= \begin{bmatrix}1 & 0& 0\\ 8& 1 & 0\\ 16*3 & 8 & 1 \end{bmatrix},$ $P^3= \begin{bmatrix}1 & 0& 0\\ 12& 1 & 0\\ 16*6 & 12 & 1 \end{bmatrix},$ $P^4= \begin{bmatrix}1 & 0& 0\\ 16& 1 & 0\\ 16*10 & 16 & 1 \end{bmatrix} \\$

Here we see some patterns: 4,8,12,16 is an AP with with difference 4 and nth term of 1,3,6,10 is sum of first n terms, hence:

$P^N= \begin{bmatrix}1 & 0& 0\\ 4N& 1 & 0\\ 16 \dfrac{n(n+1)}{2} &4N & 1 \end{bmatrix} \\$ .

$\sum_{N=1}^{a_1} P^N =\sum_{N=1}^{a_1} \begin{bmatrix}1 & 0& 0\\ 4N& 1 & 0\\ 16 \dfrac{n(n+1)}{2} &4N & 1 \end{bmatrix} = \begin{bmatrix}a_1 & 0& 0\\ 4\dfrac{a_1(a_1+1)}{2}& a_1 & 0\\ 16 \dfrac{a_1(a_1+1)(a_1+2)}{3*2} &4\dfrac{a_1(a_1+1)}{2} & a_1 \end{bmatrix} \\$

$\sum_{a_1=1}^{a_2} \sum_{N=1}^{a_1} P^N =\sum_{a_1=1}^{a_2} \begin{bmatrix}a_1 & 0& 0\\ 4\dfrac{a_1(a_1+1)}{2}& a_1 & 0\\ 16 \dfrac{a_1(a_1+1)(a_1+2)}{3*2} &4\dfrac{a_1(a_1+1)}{2} & a_1 \end{bmatrix} = \begin{bmatrix}\dfrac{a_2(a_2+1)}{2} & 0& 0\\ 4\dfrac{a_2(a_2+1)(a_2+2)}{3*2}& \dfrac{a_2(a_2+1)}{2} & 0\\ 16 \dfrac{a_2(a_2+1)(a_2+2)(a_2+3)}{4*3*2} &4\dfrac{a_2(a_2+1)(a_2+2)}{3*2} & \dfrac{a_2(a_2+1)}{2}\end{bmatrix} \\$

Here we see the below pattern (which can be proved by math induction here ):

$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$

$\therefore Q = \sum_{a_{18}=1}^{M}\sum_{a_{17}=1}^{a_{18}}...\sum_{N=1}^{a_1} P^N = \begin{bmatrix}\dfrac{M(M+1)...(M+18)}{19!} & 0& 0\\ 4\dfrac{M(M+1)(M+19)}{20!}& \dfrac{M(M+1)...(M+18)}{19!} & 0\\ 16 \dfrac{M(M+1)...(M+20)}{21!} &4\dfrac{M(M+1)(M+19)}{20!} & \dfrac{M(M+1)...(M+18)}{19!}\end{bmatrix} \\$

Let $q_{11} = \alpha = \dfrac{M(M+1)...(M+18)}{19!}, \text{ and } \alpha > 0 \text{ since } M > 0$ then $Q = \begin{bmatrix}\alpha & 0& 0\\ 4\alpha\dfrac{M+19}{20}& \alpha & 0\\ 16 \alpha \dfrac{(M+19)(M+20)}{21*20} &4\alpha\dfrac{M+19}{20} & \alpha\end{bmatrix} \\$

We are given $q_{11}^2 - q_{21}^2 + q_{31}q_{33} =0 \implies = \alpha^2 - (4\alpha\dfrac{M+19}{20})^2 + 16 \alpha^2 \dfrac{(M+19)(M+20)}{21*20} =0 \\$

Since $\alpha \neq 0$ we can divide by $\alpha^2$ to get the relation $1 - (4\dfrac{M+19}{20})^2 + 16 \dfrac{(M+19)(M+20)}{21*20} =0\\$

Solving the quadratic we get $M=16,-34, M>0 \implies M =16\\$

$\therefore 7 \left(\dfrac{q_{31}+q_{32}}{q_{21}} \right) = 7 \left(\dfrac{16 \alpha\dfrac{(M+19)(M+20)}{21*20} +4\alpha\dfrac{M+19}{20}}{4\alpha\dfrac{M+19}{20}} \right) = 7*\dfrac{55}{7} = 55$