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How to solve linear eq's in 3 variables for consistency or inconsistency?

matrix method would be appreciated.

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2 Answers

The following applies to systems of n linear equations in n variables for all n ≥ 1

In the system Ax=b (A is a matrix, b is a vector) first check the determinant as for any inconsistent system, we find that |A|=0.

Now say that |A|= 0. Then there is a vector c ≠ 0 such that cA = 0, that is c1r1 + ... + cnrn = 0 (linear combination)where c = [c1 ... cn] is a row vector and r1, ...rn are rows of A. This vector is not hard to find. Once you have this vector, the system is consistent cb = 0 for all valid choices of c and inconsistent if there is a valid choice for c such that cb ≠ 0.

In the simple case given by Robert, we have

      [  1      1      1  ] = r1                             [ 1 ]

A= [  1      1      1  ] = r2    and    b = [ 2 ]

      [  1      1      1  ] = r3                    [ 3 ]

since r1 + r2 - 2r3 = 0

we can take c = [1  1  -2] so that cA = 0.

But cb = 1*1 + 1*2 - 2*3 = -3 ≠ 0.

Therefore the system is inconsistent.

Although in Robert's example, inconsistency is trivial as all equations have the same left hand side expressions but have distinct constants on their right hand sides so that the planes are parallel and any triple (x,y,z) satisfies at most one equation.

Note, the planes don't necessarily need to be parallel at all for inconsistency for n=3 case. Consider three equations for planes containing rectangular sides of a triangular prism. The three planes intersect in pairs along lines, but those three lines are parallel and the system is inconsistent. The same discussion applies for n>3.


Consistency means you have meanful solutions. If det[matrix of coefficients] ≠0, then the solution is unique. But if det[matrix of coefficients] = 0, you have to do further work to see whether it is consistent. In general, if there are no at least two parallel planes, then the solution is consistent.


Inconsistency means there are at least two parallel planes. Therefore, there is no solution. Necessary condition: det[matrix of coefficients] = 0

For example: x +y +z = 1, x +y +z = 2 and x +y +z = 3 are inconsistent.