Test of consistency

CONSISTENT SYSTEM:

If the system of equations has one or more solutions, then it is said to be a consistent system of equations, otherwise it is an inconsistent system of equations.

For example, the system of linear equations

2x ± 3y = 5, 4x ± 6y = 10

Is consistent, because x=1, y=1 and x=2, y=1/3 are solutions of it.

However, the system of linear equations

2x ± 3y = 5, 4x ± 6y = 9

Is inconsistent, because there is no set of values of x, y which satisfy the two equations simultaneously.

CONSISTENCY AND INCONSISTENCY:

Consider the system of simultaneous linear equations given by

AX = B, where |A|= 0 … (i)

• (adj A) (AX) = (adj A) B
• ((adj A) A) X = (adj A) B
• (|A| In) X = (adj A) B
• |A| (In X) = (adj A) B
• |A| X = (adj A) B … (ii)

Now, two cases arise:

CASE I: When (adj A) B = 0.

In this case, equation (ii) is true for every value of X. So, we say that the system of equations given in (i) is consistent and it has infinitely many solution.

CASE II: When (adj A) B ≠ 0

In this case, equation (ii) does not exist because its LHS is zero and RHS is non-zero. So, we say that the given system has no solution i.e. it is an inconsistent system of equations.

ALGORITHM:

STEP I: Obtain the system of equations and express it in the matrix form AX=B.

STEP II: Find lAl. If |A| ≠ 0, then the given system of equations is consistent with unique solution given by X = A-1B.

If |A| = 0, then compute (adj A) B.

If (adj A) B = 0, then the given system of equations is consistent with infinitely many solutions.

In order to find these infinitely many solutions, replace one of the variable by some real number. This will reduce the number of variable by one. Now, take any two out of the three equations and solve them by matrix method.

If (adj A) B ≠ 0, then the given system is inconsistent i.e. it has no solution.