Key Concepts and Formulas

1. System of Linear Equations in Matrix Form: A system of linear equations can be represented as $AX=B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants.
2. Determinant of Coefficient Matrix ($\Delta$): The determinant of the coefficient matrix $A$ is denoted as $\Delta = \det(A)$.
3. Cramer's Rule Conditions for Solutions:
   • Unique Solution: If $\Delta \neq 0$, the system has a unique solution.
   • No Solution: If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ (determinants formed by replacing columns of $A$ with $B$) is non-zero, the system is inconsistent and has no solution.
   • Infinitely Many Solutions: If $\Delta = 0$ and all of $\Delta_x = \Delta_y = \Delta_z = 0$, the system is consistent and has infinitely many solutions.

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Step-by-Step Solution

Step 1: Express the system of equations in matrix form. We are given the system: $2 x+4 y+2 a z=b$ $x+2 y+3 z=4$ $2 x-5 y+2 z=8$ This can be written as $AX=B$, where: $A = \begin{bmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} b \\ 4 \\ 8 \end{bmatrix}$

Step 2: Calculate the determinant of the coefficient matrix, $\Delta$. This determinant is crucial for determining the nature of solutions. $\Delta = \det(A) = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{vmatrix}$ Expanding along the first row: $\Delta = 2 \begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} - 4 \begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} + 2a \begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix}$ $\Delta = 2((2)(2) - (3)(-5)) - 4((1)(2) - (3)(2)) + 2a((1)(-5) - (2)(2))$ $\Delta = 2(4 + 15) - 4(2 - 6) + 2a(-5 - 4)$ $\Delta = 2(19) - 4(-4) + 2a(-9)$ $\Delta = 38 + 16 - 18a$ $\Delta = 54 - 18a$ Factoring out $-18$: $\Delta = -18(a - 3)$

Step 3: Analyze conditions for a unique solution. A unique solution exists if and only if $\Delta \neq 0$. From $\Delta = -18(a - 3)$, we see that $\Delta \neq 0$ if $a \neq 3$. Let's check options (C) and (D):

• (C) It has unique solution if $a=b=8$. Here $a=8$. Since $a=8 \neq 3$, $\Delta = -18(8-3) = -18(5) = -90 \neq 0$. Thus, a unique solution exists. This statement is correct.
• (D) It has unique solution if $a=b=6$. Here $a=6$. Since $a=6 \neq 3$, $\Delta = -18(6-3) = -18(3) = -54 \neq 0$. Thus, a unique solution exists. This statement is correct.

Step 4: Analyze conditions for infinitely many solutions or no solution (when $\Delta = 0$). If $\Delta = 0$, then $-18(a-3) = 0$, which implies $a=3$. When $a=3$, we need to calculate $\Delta_x, \Delta_y, \Delta_z$ to determine if there are infinitely many solutions or no solution. Let's calculate $\Delta_z$ first, as it's often simpler to start with and can rule out options quickly. $\Delta_z = \begin{vmatrix} 2 & 4 & b \\ 1 & 2 & 4 \\ 2 & -5 & 8 \end{vmatrix}$ Expanding along the first row: $\Delta_z = 2 \begin{vmatrix} 2 & 4 \\ -5 & 8 \end{vmatrix} - 4 \begin{vmatrix} 1 & 4 \\ 2 & 8 \end{vmatrix} + b \begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix}$ $\Delta_z = 2((2)(8) - (4)(-5)) - 4((1)(8) - (4)(2)) + b((1)(-5) - (2)(2))$ $\Delta_z = 2(16 + 20) - 4(8 - 8) + b(-5 - 4)$ $\Delta_z = 2(36) - 4(0) + b(-9)$ $\Delta_z = 72 - 9b$ Factoring out $-9$: $\Delta_z = -9(b - 8)$

Now, let's evaluate options (A) and (B) where $a=3$:

• (A) It has infinitely many solutions if $a=3, b=8$.

  • For $a=3$, we have $\Delta = 0$.
  • For $b=8$, we have $\Delta_z = -9(8-8) = 0$.
  • To confirm infinitely many solutions, we must also check $\Delta_x$ and $\Delta_y$ for $a=3, b=8$.
    • $\Delta_x = \begin{vmatrix} b & 4 & 2a \\ 4 & 2 & 3 \\ 8 & -5 & 2 \end{vmatrix} = \begin{vmatrix} 8 & 4 & 6 \\ 4 & 2 & 3 \\ 8 & -5 & 2 \end{vmatrix}$
    • Notice that $R_1 = 2R_2$ (i.e., $(8,4,6) = 2(4,2,3)$). When one row is a scalar multiple of another row, the determinant is 0. So, $\Delta_x = 0$.
    • $\Delta_y = \begin{vmatrix} 2 & b & 2a \\ 1 & 4 & 3 \\ 2 & 8 & 2 \end{vmatrix} = \begin{vmatrix} 2 & 8 & 6 \\ 1 & 4 & 3 \\ 2 & 8 & 2 \end{vmatrix}$
    • Notice that $C_2 = 4C_1$ in the first two columns (i.e., $(8,4,8)^T = 4(2,1,2)^T$). When one column is a scalar multiple of another column, the determinant is 0. So, $\Delta_y = 0$.
  • Since $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$ when $a=3$ and $b=8$, the system has infinitely many solutions. This statement is correct.

• (B) It has infinitely many solutions if $a=3, b=6$.

  • For $a=3$, we have $\Delta = 0$.
  • For $b=6$, we have $\Delta_z = -9(6-8) = -9(-2) = 18$.
  • Since $\Delta = 0$ but $\Delta_z \neq 0$, the system has no solution.
  • Therefore, the statement that it has infinitely many solutions if $a=3, b=6$ is NOT correct.

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Common Mistakes & Tips

• Determinant Calculation Errors: Be very careful with signs and arithmetic when calculating determinants. Double-check your work, especially for 3x3 matrices.
• Misinterpreting Cramer's Rule: Remember the precise conditions for unique, no, and infinitely many solutions. $\Delta=0$ alone is not enough to conclude no solution or infinitely many solutions; you must check $\Delta_x, \Delta_y, \Delta_z$.
• Row/Column Operations: Using row/column operations to simplify determinants (e.g., creating zeros) can reduce calculation errors, but ensure you apply them correctly. For instance, if two rows or columns are proportional, the determinant is zero.

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Summary

We analyzed the given system of linear equations using the determinant method. We first calculated the determinant of the coefficient matrix, $\Delta = -18(a-3)$. For unique solutions, $\Delta \neq 0$, which means $a \neq 3$. Options (C) and (D) fall into this category and are correct. For $a=3$, $\Delta=0$, so we proceeded to calculate $\Delta_z = -9(b-8)$. For option (A) where $a=3, b=8$, we found $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$, indicating infinitely many solutions, making statement (A) correct. For option (B) where $a=3, b=6$, we found $\Delta=0$ but $\Delta_z=18 \neq 0$, which indicates no solution. Therefore, the statement in option (B) is incorrect.

The final answer is $\boxed{B}$