       Re: NSolve fails where Solve succeeds!

• To: mathgroup at smc.vnet.net
• Subject: [mg43625] Re: NSolve fails where Solve succeeds!
• From: "Dana DeLouis" <delouis at bellsouth.net>
• Date: Fri, 26 Sep 2003 04:45:38 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```It appears to me that in Mathematica 5, it is setting each element of
the list equal to the right hand side.

Solve[{{2*a - 3, 4*b - 1}, {3*c - 3, 5*d - 1}} == 0]

{{a -> 3/2, b -> 1/4, c -> 1, d -> 1/5}}

It likes your example.  All value are equal to zero.
Solve[{{a,b},{c,d}}==0]

However, given the above, I don't see why the following fails.  Here, it
does not return all variable as equal to 1.
Solve[{{a,b},{c,d}}==1]

It says it is "Not a well-formed equation."  Hmm?

It doesn't like Nsolve like you mentioned.
NSolve[{{2*a - 3, 4*b - 1},{3*c - 3, 5*d - 1}} == 0]

RowBox[{\(Solve::"elist"\), ":", "\<\"List encountered during
<snip>

It appears to like Reduce in some situations, similar to Solve:

It liked this:
Reduce[{{2*a - 3, 4*b - 1},  {3*c - 3, 5*d - 1}} == 0]

d == 1/5 && c == 1 && b == 1/4 && a == 3/2

It like this:
Reduce[{{a, b}, {c, d}} == 0]
d == 0 && c == 0 && b == 0 && a == 0

It did not like this:

Reduce[{{a, b}, {c, d}} == 1]
"....is not a quantified system of equations and inequalities. "

It appears to me that Solve and Reduce will work on each element only if
the right hand side ==0.

Suppose we have this list:
equ = {{2*a - 3, 4*b - 1}, {3*c - 3, 5*d - 1}}

Suppose we want to solve this list equal to 1.  This will not work
Solve[equ == 1]

However, subtracting 1 from each side will solve for the equations:
Solve[equ - 1 == 0]

{{a -> 2, b -> 1/2, c -> 4/3,  d -> 2/5}}

```

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