Given that p+q+r=0, solve the following equations for x: 

p-x r q
r q-x p
q p r-x

the entire matrix should be set equal to 0. (I guess this means that its determinant is 0)

asked Mar 19 in Calculus Answers by anonymous

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1 Answer

 

If we assume that, not only is the determinant zero, but that the components of the determinant give rise to three equations that equate to zero, then we have four equations and four unknowns. This implies we have sufficient information to find all four unknowns.

The determinant is:

(p-x)[(q-x)(r-x)-p²]-r[r(r-x)-pq]+q[pr-q(q-x)]=0

If each of the square-bracketed expressions is zero then the determinant will also be zero. So:-

(q-x)(r-x)-p²=0 ①

r(r-x)-pq=0 ② 

pr-q(q-x)=0 ③ 

Also r=-(p+q). So ③ becomes -p(p+q)-q(q-x)=0, -p²-pq-q²+qx=0, x=(p²+pq+q²)/q.

And ② becomes r²-rx-pq=0, (p+q)²+(p+q)(p²+pq+q²)/q-pq=0.

p²+pq+q²+(p+q)(p²+pq+q²)/q=0=(p²+pq+q²)(1+(p+q)/q).

The quadratic has complex roots, so choose 1+(p+q)/q=0 as the better option.

This means p+2q=0 and p=-2q. Therefore x=(4q²-2q²+q²)/q=3q.

Finally r=-(p+q)=q.

Now substitute these values for p, r and x into each equation:

(-2q)(-2q)-4q²=0 is true

q(-2q)+2q²=0 is true

-2q²-q(-2q)=0 is true

There are many solutions because there is no unique value for q.

So q is arbitrary, and p=-2q, r=q and x=3q. So the matrix can be rewritten:

-5q q q
q -2q -2q
q -2q -2q

It can be easily seen that the determinant is zero. The common factor can be taken outside the matrix and the matrix simply becomes the coefficients -5, 1 and -2.

 

answered Mar 21 by Rod Top Rated User (571,200 points)

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