For the system of linear equations
2x+4y+2az=b
x+2y+3z=4
2x-5y+2z=8
which of the following is NOT correct ?
2x+4y+2az=b
x+2y+3z=4
2x-5y+2z=8
which of the following is NOT correct ?
- It has unique solution if a = b = 6
- It has unique solution if a = b = 8
- It has infinitely many solution if a = 3, b = 8
- It has infinitely many solutions if a = 3, b = 6
The Correct Option is D
Solution and Explanation
Solution:
Let the coefficient matrix be
\[
A = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{vmatrix}, \quad \Delta = \det(A).
\]
Expanding:
\[
\Delta = 2\begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} - 4\begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} + 2a\begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix}.
\]
Evaluating each minor:
\[
\begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} = 4 + 15 = 19, \quad \begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} = 1 \cdot 2 - 3 \cdot 2 = 2 - 6 = -4, \quad \begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix} = (1)(-5) - (2)(2) = -5 - 4 = -9.
\]
So
\[
\Delta = 2 \cdot 19 - 4 \cdot (-4) + 2a \cdot (-9) = 38 + 16 - 18a = 54 - 18a = 18(3 - a).
\]
For a unique solution, we need \(\Delta \neq 0\), hence \(a \neq 3\).
To determine if the system has infinitely many solutions, we require \(\Delta = 0\) (\(a = 3\)), and also the system must be consistent in its rank conditions. Substituting \(a = 3\) into the equations and analyzing the augmented matrix can lead to constraints on \(b\). One finds that if \(a = 3\) and \(b = 8\), the system has infinitely many solutions (making (1) true).
Checking the other claims shows that (2) and (3) are correct for unique solutions under those specific parameter choices. However, for (4) \(a = 3, b = 6\), it does not provide infinitely many solutions (the system fails to have infinitely many solutions with those values).
Hence, statement (4) is not correct.
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