1. Key Concept: Conditions for Infinitely Many Solutions

For a system of linear equations in the form: $\begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned}$ to have infinitely many solutions, it must satisfy two crucial conditions based on Cramer's Rule:

1. The determinant of the coefficient matrix, $D$, must be zero.
2. All the determinants $D_x$, $D_y$, and $D_z$ (formed by replacing the respective coefficient columns with the constant terms) must also be zero.

Mathematically, this means: $D = \left| {\begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix}} \right| = 0 \quad \text{AND} \quad D_x = \left| {\begin{matrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{matrix}} \right| = 0 \quad \text{AND} \quad D_y = \left| {\begin{matrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{matrix}} \right| = 0 \quad \text{AND} \quad D_z = \left| {\begin{matrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{matrix}} \right| = 0$ If $D=0$ but at least one of $D_x, D_y, D_z$ is non-zero, the system has no solution (inconsistent).

2. Step-by-Step Solution

Given the system of equations:

1. $x - 2y + 3z = 9$
2. $2x + y + z = b$
3. $x - 7y + az = 24$

Step 1: Calculate the Determinant of the Coefficient Matrix (D) and set it to zero.

First, we form the coefficient matrix and calculate its determinant, $D$. This is the primary condition for the system to either have no solution or infinitely many solutions. If $D \ne 0$, there would be a unique solution.

The coefficient matrix is: $A = \left( {\begin{matrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{matrix}} \right)$ Setting $D = \det(A) = 0$: $D = \left| {\begin{matrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{matrix}} \right| = 0$ Expand the determinant along the first row: $1 \cdot ((1)(a) - (1)(-7)) - (-2) \cdot ((2)(a) - (1)(1)) + 3 \cdot ((2)(-7) - (1)(1)) = 0$ $1(a + 7) + 2(2a - 1) + 3(-14 - 1) = 0$ $a + 7 + 4a - 2 - 45 = 0$ Combine like terms: $5a - 40 = 0$ $5a = 40$ $a = 8$ So, we have found the value of $a$.

Step 2: Calculate $D_x$ (or $D_1$) and set it to zero.

Now that we have $a=8$ (which makes $D=0$), we need to ensure that the system is consistent and has infinitely many solutions, not no solution. For this, we must check that $D_x$, $D_y$, and $D_z$ are also zero. It's usually sufficient to use one or two of these conditions to find the remaining unknown variables. Let's use $D_x$.

$D_x$ is formed by replacing the first column of the coefficient matrix with the constant terms (9, b, 24): $D_x = \left| {\begin{matrix} 9 & -2 & 3 \\ b & 1 & 1 \\ 24 & -7 & a \end{matrix}} \right|$ Substitute $a=8$ into $D_x$: $D_x = \left| {\begin{matrix} 9 & -2 & 3 \\ b & 1 & 1 \\ 24 & -7 & 8 \end{matrix}} \right|$ Set $D_x = 0$ and expand the determinant along the first row: $9 \cdot ((1)(8) - (1)(-7)) - (-2) \cdot ((b)(8) - (1)(24)) + 3 \cdot ((b)(-7) - (1)(24)) = 0$ $9(8 + 7) + 2(8b - 24) + 3(-7b - 24) = 0$ $9(15) + 16b - 48 - 21b - 72 = 0$ $135 + 16b - 48 - 21b - 72 = 0$ Combine constant terms and 'b' terms: $(135 - 48 - 72) + (16b - 21b) = 0$ $(135 - 120) - 5b = 0$ $15 - 5b = 0$ $5b = 15$ $b = 3$ We have now found $b=3$.

Step 3: Calculate the required value.

The question asks for the value of $a - b$. $a - b = 8 - 3 = 5$

3. Common Mistakes & Tips

• Incomplete Conditions: A very common mistake is to only check $D=0$ and assume infinitely many solutions. Remember that $D=0$ can also lead to no solution (inconsistent system) if any of $D_x, D_y, D_z$ are non-zero. Always check the other determinants for consistency.
• Determinant Calculation Errors: Be extremely careful with signs and arithmetic during determinant expansion. A single sign error can lead to a completely wrong answer.
• Alternative Methods: While Cramer's rule conditions are generally robust, for some problems, row reduction (Gaussian elimination) can be an alternative, especially if the coefficients are simple. For infinitely many solutions, row reduction would lead to a row of zeros (0 = 0) after reaching row echelon form.
• Verification (Optional but Recommended): Ideally, you should also calculate $D_y$ and $D_z$ with $a=8$ and $b=3$ and verify that they are indeed zero. This confirms that the system is truly consistent with infinitely many solutions. For example, $D_y = \left| {\begin{matrix} 1 & 9 & 3 \\ 2 & 3 & 1 \\ 1 & 24 & 8 \end{matrix}} \right| = 1(24-24) - 9(16-1) + 3(48-3) = 0 - 9(15) + 3(45) = -135 + 135 = 0$. This confirms our values.

4. Summary & Key Takeaway

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix ($D$) must be zero, AND all the determinants $D_x$, $D_y$, $D_z$ (formed by replacing coefficient columns with constant terms) must also be zero. By applying these conditions sequentially, we found $a=8$ and $b=3$, leading to $a-b=5$.