QJEEM23Ma8357: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8357 Solution: For the system of linear equations which of the following is…

The given system of equations is :

1. $2x + 4y + 2az = b$

2. $x + 2y + 3z = 4$

3. $2x - 5y + 2z = 8$

We can write this in matrix form :

$\begin{bmatrix}2 & 4 & 2a \\1 & 2 & 3 \\2 & -5 & 2\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}b \\4 \\8\end{bmatrix}$

First, we need to find the determinant of the coefficient matrix, which we'll call Δ. The coefficient matrix is :

$\begin{bmatrix}2 & 4 & 2a \\1 & 2 & 3 \\2 & -5 & 2\end{bmatrix}$

We find its determinant for 3 $\times$ 3 matrices :

$\Delta = 2(2)(2) + 4(3)(2) + 2a(1)(-5) - 2a(2)(2) - 4(1)(2) - 2(3)(-5)$

$\Delta = 8 + 24 - 10a - 8a - 8 + 30$

$\Delta = -18a + 54$

$\Delta = -18(a - 3)$

Next, we substitute the third column of our matrix with the column of constants (b, 4, 8) and calculate the determinant Δ₃ :

$\begin{bmatrix}2 & 4 & b \\1 & 2 & 4 \\2 & -5 & 8\end{bmatrix}$

We find its determinant :

$\Delta_3 = 2(2)(8) + 4(4)(2) + b(1)(-5) - b(2)(2) - 4(1)(8) - 2(4)(-5)$

$\Delta_3 = 32 + 32 - 5b - 4b - 32 + 40$

$\Delta_3 = -9b + 72$

$\Delta_3 = 9(8 - b)$

Now, let's analyze the options :

1. For a=3 and b=8, we have Δ = 0 and Δ₃ = 0, which indicates an infinite number of solutions.

2. For a=3 and b=6, we have Δ = 0 and Δ₃ ≠ 0, which indicates no solution.

3. For a=8 and b=8, we have Δ ≠ 0, which indicates a unique solution.

4. For a=6 and b=6, we have Δ ≠ 0, which indicates a unique solution.

Therefore, the statement that is NOT correct is Option B: "It has infinitely many solutions if a=3, b=6", because in this case the system actually has no solution.