Question

[img]https://upload.cc/i1/2020/03/12/Bl1aVO.jpg[/img]



R is 4

Answer 1

(a)

$\begin{vmatrix} 1 & pq & r^2 \\ 1 & qr & p^2\\ 1 & pr & q^2 \end{vmatrix}= (p-q)(q-r)(p-r)(p+q+r)$

$\begin{vmatrix} 1 & pq & r^2 \\ 1 & qr & p^2\\ 1 & pr & q^2 \end{vmatrix}=\begin{vmatrix} qr & p^2 \\ pr & q^2 \\ \end{vmatrix}-\begin{vmatrix} pq & r^2\\ pr & q^2 \end{vmatrix}+\begin{vmatrix} pq & r^2\\ qr & p^2 \end{vmatrix}=\\ =q^3r-p^3r-pq^3+pr^3+p^3q-qr^3=(q^3r-p^3r)+(p^3q-pq^3)+(pr^3-qr^3)=\\ =r(q^3-p^3)+pq(p^2-q^2)+r^3(p-q)=-r(p-q)(p^2+pq+q^2)+pq(p-q)(p+q)+\\ +r^3(p-q)=(p-q)(-r(p^2+pq+q^2)+pq(p+q)+r3)=(p-q)(-p^2r-pqr-q^2r+\\ +p^2q+pq^2+r^3)=(p-q)((p^2q-p^2r)-(q^2r-r^3)+(pq^2-pqr))=\\ =(p-q)(p^2(q-r)-r(q^2-r^2)+pq(q-r))=(p-q)(p^2(q-r)-r(q-r)(q+r)+\\ +pq(q-r))=(p-q)(q-r)(p^2-r(q+r)+pq)=(p-q)(q-r)(p^2-rq-r^2+pq)=\\ =(p-q)(q-r)((p^2-r^2)+(pq-rq))=(p-q)(q-r)((p-r)(p+r)+q(p-r))=\\ =(p-q)(q-r)(p-r)(p+q+r)$

$(b)\begin{pmatrix} 1 & (R+1)x & (R+3)^2\\ 1 & (R+3)x & (R+1)^2\\ 1 & (R+1)(R+3) & x^2 \end{pmatrix} is \ a\ singular \ matrix \rArr$

$\begin{vmatrix} 1 & (R+1)x & (R+3)^2\\ 1 & (R+3)x & (R+1)^2\\ 1 & (R+1)(R+3) & x^2 \end{vmatrix}=0$

$\begin{vmatrix} 1 & (R+1)x & (R+3)^2\\ 1 & (R+3)x & (R+1)^2\\ 1 & (R+1)(R+3) & x^2 \end{vmatrix}=((R+1)-x)((R+3)-x)((R+3)-(R+1))\\ ((R+1)+(R+3)+x)=2(R+1-x)(R+3-x)(2R+4+x)=0\\ x=R+1 \ or\\ x=R+3 \ or\\ x=-2R-4$