Question

If \[p\ne 0,q\ne 0\] and \[\left| \begin{matrix}
   p & q & p\alpha +q \\
   q & r & q\alpha +r \\
   p\alpha +q & q\alpha +r & 0 \\
\end{matrix} \right|=0\], then using property of determinants prove that either p, q, r are in G.P. or \[\alpha \] is a root of the equation \[p{{x}^{2}}+2qx+r=0\].

Answer 1

Hint: Don’t expand the determinant directly. Use the following property to make the given determinant in a simpler form as \[{{R}_{3}}\to {{R}_{3}}-(\alpha {{R}_{1}}+{{R}_{2}})\] And hence, expand the given determinant along the row 3. And use the following results to prove the given statement as: -

Complete step-by-step solution -
a, b, c are in G.P, if \[{{b}^{2}}=ac\]
α will be a root of \[f(x)\] if \[f(\alpha)\]=0
If, xy = 0. Then x=0 or y=0
Given expression in the problem is
\[\left| \begin{matrix}
   p & q & p\alpha +q \\
   q & r & q\alpha +r \\
   p\alpha +q & q\alpha +r & 0 \\
\end{matrix} \right|=0\] ……… (i)
Where, \[p\ne 0,q\ne 0\] and hence, we need to prove that either p, q, r will be in G.P or α is a root of\[p{{x}^{2}}+2qx+r=0\].
So, let us solve the determinant of equation (i) and equate it to 0, where, we have to use properties of determinant to get the value of the determinant.
Now, we can observe the elements of the given determinant and get that if, we multiple the elements of first row by ‘α’ and add it to the corresponding element of second row, then if we subtract the result from the corresponding element of third row, we get that first two elements of third row, will become 0. In other words, if we subtract \[p\alpha +q\] and \[q\alpha +r\] from first and second element of third row, they will become 0 individually. So, we can apply the property with the determinant as \[{{R}_{3}}\to {{R}_{3}}-(\alpha {{R}_{1}}+{{R}_{2}})\]…….. (ii)
So, we can re-write the equation (i) by applying the above property as
\[\left| \begin{matrix}
   p & q & p\alpha +q \\
   q & r & q\alpha +r \\
   (p\alpha +q)-(p\alpha +q) & (q\alpha +r)-(q\alpha +r) & 0-[\alpha (p\alpha +q)+(q\alpha +r)] \\
\end{matrix} \right|\]\[\] = 0
Or
\[\left| \begin{matrix}
   p & q & p\alpha +q \\
   q & r & q\alpha +r \\
   0 & 0 & -(p{{\alpha }^{2}}+q\alpha +q\alpha +r) \\
\end{matrix} \right|\] = 0
Or
\[\left| \begin{matrix}
   p & q & p\alpha +q \\
   q & r & q\alpha +r \\
   0 & 0 & -(p{{\alpha }^{2}}+2q\alpha +r) \\
\end{matrix} \right|\] = 0
Now, we can expand the determinant along row 3 to get the value of the determinant. Hence, we get the above equation as:
\[[0(q(q\alpha +r)-r(p\alpha +q))-0(p(q\alpha +r)-q(p\alpha +q))-(p{{\alpha }^{2}}+2p\alpha +r)[pr-{{q}^{2}}]]\] = 0
On simplifying the above equation further, we get
\[\begin{align}
  & 0-0-(p{{\alpha }^{2}}+2q\alpha +r)[pr-{{q}^{2}}]=0 \\
 & or \\
 & -(p{{\alpha }^{2}}+2q\alpha +r)[pr-{{q}^{2}}]=0 \\
\end{align}\]
On multiplying the above equating by -1, we can re-write the equation as
\[(p{{\alpha }^{2}}+2q\alpha +r)[pr-{{q}^{2}}]=0\]……….. (iii)
Now, as we know two or three variables present in multiplication form can be 0, if either of them is 0. So, if xy=0 then x=0 or y=0.
Now, we can observe equation (iii), we get that \[(p{{\alpha }^{2}}+2q\alpha +r)\] and \[(pr-{{q}^{2}})\] are in product form and equal to 0. So, either of them should be 0.
Hence, we get \[pr={{q}^{2}}\] or \[p{{\alpha }^{2}}+2q\alpha +r=0\]
Now, as we know, the numbers a, b, c will be in G.P, if
\[{{b}^{2}}=ac\] …… (iv)
And any polynomial \[f(x)\] has root ‘a’, if
\[f(a)\] = 0 ……. (v)
Hence, we get that p, q, r are in G.P from the relation \[{{q}^{2}}=pr\] and equation (i).
And, also we get that α will be a root of equation \[p{{\alpha }^{2}}+2q\alpha +r\] from the relation \[p{{\alpha }^{2}}+2q\alpha +r=0\] and equation (iii). So, the given statements are proved.


Note: One may directly expand the given determinant along any row or any column, but that would be a very complex approach and may not get the result which is required because factorization of the expression will be difficult. It will be a longer approach as well, so try to use the property of determinant with these kinds of questions to make the solution flexible and less time taking.
Another approach would be that we can use a different property with the given determinant as \[{{c}_{3}}\to {{c}_{3}}-(\alpha {{c}_{1}}+{{c}_{2}})\]
The above property can also be applied.