If $bc+qr=ca+rp=ab+pq=-1$ then show that $\begin{vmatrix} ap & p & a \\ bq & q & b \\ cr & r & c \\ \end{vmatrix}=0 $

Question

If $bc+qr=ca+rp=ab+pq=-1$ then show that
$\begin{vmatrix} ap & \color{red} {p} & a \\ bq & q & b \\ cr & r & c \\ \end{vmatrix}=0 $
Try

we are given that $1 + b c + q r = 0$
$ 1 + c a + p r = 0$
$1+ab+pq=0$

The determinant in the question involves a column consisting the elements ap, bq and cr So multipying (1),(2) and (3) by ap bq and cr respectively we get
$ap+abcp+apqr=0$
$b q + a b c q + b p q r = 0$
$c r + a b c r + c p q r = 0$ Since abc and pqr occur in all the three equations putting $abc= x$ and $pqr=y$ we get
$a p + p x + a y = 0$
$bq+qx+\color{red}{b} y=0$
$cr+rx+cy=0$
Now, since there are only two variable and three equations hence to have a solution one of the equation must be a linear combination of the other hence the coefficient matrix is linearly dependent i.e. the coefficient determinant is zero.
But the book I am reading has introduced the problem before it discusses solving technique of system of linear equation. So, I wonder if there is any other way to look at this problem. Thanks in advance.

Answer 1

By direct computations, \begin{align*} \begin{vmatrix} ap & p & a \\ bq & q & b \\ cr & r & c \\ \end{vmatrix} &= apqc-aprb-bqpc+bqra+crpb-crqa\\ &=ac(pq-rq)+ab(rq-rp)+bc(rp-pq)\\ &=(-1-rp)(pq-rq)+(-1-pq)(rq-rp)+(-1-rq)(pr-pq)\\ &=rq-pq+rp-rq+pq-pr +rprq-rppq+rppq-rqqp+rqpq-rqpr\\ &=0 \end{align*}

Answer 2

I found a counterexample. Let $$ (a,b,c,p,q,r)=(1,1,-1,-2,1,0). $$ Then the equations are satisfied, i.e., we have $$bc+qr=ca+rp=ab+pq=-1,$$ but the determinant of the matrix is $3$. Indeed, $$ \det \begin{pmatrix} -2 & 1 & 1 \cr 1 & 1 & 1 \cr 0 & 0 & -1\end{pmatrix} =3. $$

Edit: The corrected question can be answered by a direct computation. No need of further linear algebra. The formula for the $3\times 3$ determinant is also called Rule of Sarrus.