1. Key Concepts: Conditions for Solutions of a System of Linear Equations

For a system of linear equations in the form $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix, we can analyze the nature of its solutions using determinants.

Let $A$ be the coefficient matrix: $A = \begin{pmatrix} -1 & 1 & 2 \\ 3 & -a & 5 \\ 2 & -2 & -a \end{pmatrix}$ Let $B$ be the constant matrix: $B = \begin{pmatrix} 0 \\ 1 \\ 7 \end{pmatrix}$

We define the following determinants:

• $\Delta = \det(A)$
• $\Delta_x$: Determinant of the matrix formed by replacing the first column of $A$ with $B$.
• $\Delta_y$: Determinant of the matrix formed by replacing the second column of $A$ with $B$.
• $\Delta_z$: Determinant of the matrix formed by replacing the third column of $A$ with $B$.

The conditions for the types of solutions are:

1. Unique Solution: The system has a unique solution if and only if $\Delta \neq 0$.
2. Inconsistent (No Solution): The system is inconsistent if and only if $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
3. Infinitely Many Solutions: The system has infinitely many solutions if and only if $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$.

Our goal is to find the values of $a \in \mathbb{R}$ for which the system is inconsistent ($S_1$) and for which it has infinitely many solutions ($S_2$).

2. Step-by-Step Working

Step 1: Calculate the determinant of the coefficient matrix, $\Delta$. The first step is to calculate $\Delta = \det(A)$. This determinant is crucial because its value dictates whether the system has a unique solution ($\Delta \neq 0$) or might have no solution or infinitely many solutions ($\Delta = 0$). We are interested in the latter cases for $S_1$ and $S_2$.

$\Delta = \begin{vmatrix} -1 & 1 & 2 \\ 3 & -a & 5 \\ 2 & -2 & -a \end{vmatrix}$ Expanding along the first row: $\Delta = -1 \begin{vmatrix} -a & 5 \\ -2 & -a \end{vmatrix} - 1 \begin{vmatrix} 3 & 5 \\ 2 & -a \end{vmatrix} + 2 \begin{vmatrix} 3 & -a \\ 2 & -2 \end{vmatrix}$ $\Delta = -1((-a)(-a) - 5(-2)) - 1(3(-a) - 5(2)) + 2(3(-2) - (-a)(2))$ $\Delta = -1(a^2 + 10) - 1(-3a - 10) + 2(-6 + 2a)$ $\Delta = -a^2 - 10 + 3a + 10 - 12 + 4a$ $\Delta = -a^2 + 7a - 12$

Step 2: Find values of 'a' for which $\Delta = 0$. To identify when the system might be inconsistent or have infinitely many solutions, we set $\Delta = 0$. These are the critical values of $a$ we need to investigate further.

$-a^2 + 7a - 12 = 0$ Multiplying by -1 to simplify: $a^2 - 7a + 12 = 0$ Factoring the quadratic equation: $(a-3)(a-4) = 0$ This gives us two critical values for $a$: $a=3$ and $a=4$.

Step 3: Analyze the system for each critical value of 'a'. For each value of $a$ where $\Delta = 0$, we need to calculate $\Delta_x$, $\Delta_y$, and $\Delta_z$. If at least one of these is non-zero, the system is inconsistent. If all three are zero, the system has infinitely many solutions.

Case 1: When $a=3$ First, we substitute $a=3$ into the matrix for $\Delta_x$: $\Delta_x = \begin{vmatrix} 0 & 1 & 2 \\ 1 & -3 & 5 \\ 7 & -2 & -3 \end{vmatrix}$ Expanding along the first column (this method can be less prone to sign errors when the first element is 0): $\Delta_x = 0 \begin{vmatrix} -3 & 5 \\ -2 & -3 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ -2 & -3 \end{vmatrix} + 7 \begin{vmatrix} 1 & 2 \\ -3 & 5 \end{vmatrix}$ $\Delta_x = 0 - 1(1(-3) - 2(-2)) + 7(1(5) - 2(-3))$ $\Delta_x = -1(-3 + 4) + 7(5 + 6)$ $\Delta_x = -1(1) + 7(11) = -1 + 77 = 76$ Since for $a=3$, $\Delta = 0$ and $\Delta_x = 76 \neq 0$, the system is inconsistent. Therefore, $a=3 \in S_1$. We don't need to calculate $\Delta_y$ or $\Delta_z$ for this case, as finding one non-zero $\Delta_i$ is sufficient to conclude inconsistency.

Case 2: When $a=4$ Next, we substitute $a=4$ into the matrix for $\Delta_x$: $\Delta_x = \begin{vmatrix} 0 & 1 & 2 \\ 1 & -4 & 5 \\ 7 & -2 & -4 \end{vmatrix}$ Expanding along the first column: $\Delta_x = 0 \begin{vmatrix} -4 & 5 \\ -2 & -4 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ -2 & -4 \end{vmatrix} + 7 \begin{vmatrix} 1 & 2 \\ -4 & 5 \end{vmatrix}$ $\Delta_x = 0 - 1(1(-4) - 2(-2)) + 7(1(5) - 2(-4))$ $\Delta_x = -1(-4 + 4) + 7(5 + 8)$ $\Delta_x = -1(0) + 7(13) = 0 + 91 = 91$ Since for $a=4$, $\Delta = 0$ and $\Delta_x = 91 \neq 0$, the system is inconsistent. Therefore, $a=4 \in S_1$. Again, no need to calculate $\Delta_y$ or $\Delta_z$.

Step 4: Determine the sets $S_1$ and $S_2$ and their cardinalities. Based on our analysis:

• For $a=3$, the system is inconsistent. So, $3 \in S_1$.
• For $a=4$, the system is inconsistent. So, $4 \in S_1$.
• We did not find any values of $a$ for which $\Delta=0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$. This means there are no values of $a$ for which the system has infinitely many solutions.

Therefore:

• $S_1 = \{3, 4\}$
• $S_2 = \emptyset$ (the empty set)

The number of elements in $S_1$ is $n(S_1) = 2$. The number of elements in $S_2$ is $n(S_2) = 0$.

3. Common Mistakes and Tips

• Sign Errors in Determinant Calculation: These are very common. Always double-check your calculations, especially with negative signs. Expanding along a row or column with zeros (like the first column of $\Delta_x$) can simplify calculations and reduce error.
• Incomplete Analysis when $\Delta = 0$: If $\Delta = 0$, it's crucial to calculate $\Delta_x, \Delta_y, \Delta_z$. Don't stop after just finding $\Delta=0$.
  • If any of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system is inconsistent. You don't need to check the others once you find one non-zero.
  • If all of $\Delta_x, \Delta_y, \Delta_z$ are zero, then the system has infinitely many solutions.
• Alternative Method (Gaussian Elimination): For verifying your results or if determinant calculations feel too complex, you can use Gaussian elimination (row operations on the augmented matrix) to find the rank of the coefficient matrix and the augmented matrix. This provides an alternative way to classify the system. For $a=3$ and $a=4$, using row operations also confirms inconsistency (leading to equations like $0=k$ where $k \neq 0$).

4. Summary and Key Takeaway

This problem effectively tests your understanding of the conditions for solving a system of linear equations using determinants.

1. Always start by calculating $\Delta = \det(A)$.
2. Set $\Delta = 0$ to find the critical values of the parameter (here, $a$) where the system might be inconsistent or have infinitely many solutions.
3. For each critical value, calculate $\Delta_x, \Delta_y, \Delta_z$.
4. Apply the conditions:
   • $\Delta=0$ and at least one $\Delta_i \neq 0 \implies$ Inconsistent.
   • $\Delta=0$ and all $\Delta_i = 0 \implies$ Infinitely many solutions.

In this problem, we found $\Delta = -a^2 + 7a - 12$, which is zero for $a=3$ and $a=4$. For both these values, we calculated $\Delta_x$ and found it to be non-zero ($76$ for $a=3$ and $91$ for $a=4$). This directly implies that for both $a=3$ and $a=4$, the system is inconsistent. No values of $a$ led to infinitely many solutions.

The final answer is \boxed{\text{n(S_1) = 2, n(S_2) = 0}}.