How to calculate this determinant of a $2\times 2$matrix?

Question

This matrix arises from a homework problem which our professor gave.

We need to find the determinant of this matrix.

Does there exist any simple way to find the determinant of this matrix?

$\begin{pmatrix} x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) && (1-p)(2+\frac{l}{x-n+2-l})\\ (1-q)(2+\frac{l}{x-n+2-l}) && x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l}) \end{pmatrix}$

Here $n=pq$ and $l=\phi(n)+1$.

Is there any software which can calculate this large determinant?

One of my friends got $x-n+2-l$ as a factor of this determinant.

Is there any simple way to calculate this determinant?

I am stuck.

My try:

$R_1\to R_1-R_2$ gives

$\begin{pmatrix} x-pq-p+q+2 && -x+pq+q-p-2\\ (1-q)(2+\frac{l}{x-n+2-l}) && x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l}) \end{pmatrix}$

Answer 1

In case you want to see my computation, I obtain the following determinant by a direct computation and a further factorisation: $$ \det(A)=\frac{f(p,q,x)}{((2q - 1)p - (q + x))^2}, $$

where

$$ f(p,q,x)=(2p^2q^2 + 2p^2q + 2pq^2 - 3pqx - 14pq - px + 2p - qx + 2q + x^2 + 6x + 4)(2pq - p - q - x)(pq - x - 2) $$ Here I have substituted $n=pq$ and $l=\phi(n)+1=(p-1)(q-1)+1$.

Answer 2

It helps to rewrite your matrix as $A+B$ with$$A:=\left(\begin{array}{cc} x-pq-p+3 & 1-p\\ 1-q & x-pq-q+3 \end{array}\right),\,B:=\frac{x+2-n}{x+2-n-l}\left(\begin{array}{cc} 1-q & 1-p\\ 1-q & 1-p \end{array}\right).$$Write $A=\left(\begin{array}{cc} a & c\\ b & d \end{array}\right),\,B=\left(\begin{array}{cc} e & g\\ f & h \end{array}\right)$ so$$\begin{align}\det\left(A+B\right)&=\det A+\det B+\det\left(\begin{array}{cc} a & g\\ b & h \end{array}\right)+\det\left(\begin{array}{cc} e & c\\ f & d \end{array}\right)\\&=\left(x-pq+3\right)^{2}-\left(p+q\right)\left(x-pq+3\right)-1+p+q\\&+\frac{x+2-n}{x+2-n-l}\left[\det\left(\begin{array}{cc} x-pq-p+3 & 1-p\\ 1-q & 1-p \end{array}\right)+\det\left(\begin{array}{cc} 1-q & 1-p\\ 1-q & x-pq-q+3 \end{array}\right)\right]\\&=\left(x-pq+3\right)^{2}-\left(p+q\right)\left(x-pq+3\right)-1+p+q\\&+\frac{x+2-n}{x+2-n-l}\left[\left(x-\left(p-1\right)\left(q+1\right)+1\right)\left(1-p\right)+\left(x-\left(p+1\right)\left(q-1\right)+1\right)\left(1-q\right)\right]\\&=\left(x-pq+3\right)^{2}-\left(p+q\right)\left(x-pq+3\right)-1+p+q\\&+\frac{x+2-n}{x+2-n-l}\left[\left(1+x\right)\left(2-p-q\right)+\left(p-1\right)^{2}\left(q+1\right)+\left(p+1\right)\left(q-1\right)^{2}\right].\end{align}$$

Answer 3

Call your matrix $M$ and let $s=x+2-n$. Then \begin{aligned} M&=\pmatrix{ x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) &(1-p)(2+\frac{l}{x-n+2-l})\\ (1-q)(2+\frac{l}{x-n+2-l}) &x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})}\\ &=\pmatrix{x-pq-p+3 &1-p\\ 1-q &x-pq-q+3} +\frac{x+2-n}{x+2-n-l}\pmatrix{1-q&1-p\\ 1-q&1-p}\\ &=sI+\pmatrix{1-p&1-p\\ 1-q&1-q} +\frac{s}{s-l}\pmatrix{1-q&1-p\\ 1-q&1-p}. \end{aligned} (The second equality above is borrowed from J.G.'s answer.)

If $l=2s$, then $\dfrac{s}{s-l}=-1$. Hence $$ M=\pmatrix{s+q-p&0\\ 0&s+p-q} $$ and $$ \det(M)=s^2-(p-q)^2. $$ If $l\ne2s$, let $t=\dfrac{s-l}{2s-l},\ p'=\dfrac{1-p}{t}$ and $q'=\dfrac{1-q}{t}$. Then $$ M=\pmatrix{s+tp'+(1-t)q'&p'\\ q'&s+tq'+(1-t)p'} $$ and hence $$ \det(M)=\left(s+tp'+(1-t)q'\right)\left(s+tq'+(1-t)p'\right)-p'q'. $$