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given that $q^2- pr<0, p>0$ then evaluate \begin{vmatrix} p &q &px+qy\\ q &r &qx+ry\\ px+qy &qx+ry &0 \end{vmatrix}

I've tried this: \begin{vmatrix} p &q &px+qy\\ q &r &qx+ry\\ px+qy &qx+ry &0 \end{vmatrix} then, i modified the rows as $R_3=R_3-(xR_1+yR_2)$ to get \begin{vmatrix} p &q &px+qy\\ q &r &qx+ry\\ 0 &0 &px^2+2qxy+ry^2 \end{vmatrix} then I've simplified it as $(px^2+2qxy+ry^2)(pr-q^2)$, but i cannot do further. can you please give me hints?

  • If the matrix is $(a_{ij});\space i,j=1,2,3$ then the determinant is given by the expression $$a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-( a_{13}a_{22}a_{31}+a_{12}a_{21}a_{33}+a_{11}a_{23}a_{32})$$ Can you learned from this a mnemonic? – Piquito Aug 03 '16 at 02:32
  • I didn't quite get what you're trying to convey via the mnemonic. please explain. –  Aug 03 '16 at 05:14
  • Look at the indices and think at the cycle $1\to 2\to 3$. For the positive terms you have a behavior in $ij$ and another for the negative terms. It is just to remember and make directly the calculation. – Piquito Aug 04 '16 at 12:37
  • @Piquito It's not always best to use that formula for $3\times 3$ determinants, and, in fact, I think Tara's use of row reductions took much less work overall. Even though it doesn't say it, the actual question wasn't about how to compute the determinant, but how to determine whether it's positive, negative, or zero, and the stated simplified determinant is correct. – Kyle Miller Aug 06 '16 at 03:52

2 Answers2

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Just a typo, but it seems that you dropped a minus sign since you took $R_3 = xR_1 + yR_2 - R_3$, so you essentially replaced $R_3$ with its negative (which negates the determinant) then added multiples of other rows to it (which does not change the determinant).

Besides that, I don't see how you can simplify it any more. It seems to me that the conditions imposed on $p, q,$ and $r$ just make your answer have a positive coefficient for $x$.

user357980
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  • I have edited it.Please check. Also they've asked me to find out a value ( whether positive or negative or zero). So it's kinda confusing. –  Aug 03 '16 at 02:23
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First, $pr-q^2$ is positive. What about $px^2+2qxy+ry^2$? This is the quadratic form $\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}p&q\\q&r\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$. Whether this is a positive quantity comes down to whether the eigenvalues of the matrix are positive (and whether $x,y$ are nonzero). The eigenvalues are the roots of $\lambda^2-(p+r)\lambda-(q^2-pr)$, which are $\frac{1}{2}(p+r\pm\sqrt{p^2+4q^2-2pr+r^2})$. The expression $p^2+4q^2-2pr+r^2$ can be written as $(p+r)^2+4(q^2-pr)<(p+r)^2$, hence the roots are both between $0$ and $p+r$ (exclusive). The condition $p>0$ and $q^2<pr$ together imply that $r>0$ as well, so the roots are positive.

Thus, the determinant of the matrix is nonnegative, and positive if and only if $x,y$ are not both $0$.

Alternatively, instead of examining eigenvalues of the quadratic form, we can use the quadratic formula along with the intermediate value theorem. Let's see for which $x,y$ the factor $px^2+2qxy+ry^2$ is $0$. The quadratic formula gives $x=\frac{-2qy\pm\sqrt{4q^2y^2-4pry^2}}{2p}=y\frac{-q\pm \sqrt{q^2-pr}}{p}$. The discriminant is negative, so $y$ must be $0$, and so $x$ must be $0$ as well. Since the substitution $(x,y)=(1,0)$ gives $px^2+2qxy+ry^2=p>0$, the factor is always non-negative, and positive when $(x,y)\neq (0,0)$. (This is because if it were ever negative, a path between $(1,0)$ and the point where it is negative would contain a point where the factor is $0$, by the intermediate value theorem.)

Kyle Miller
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