Key Concepts and Formulas

• System of Linear Equations: A system of $n$ linear equations in $n$ variables can be represented in matrix form as $AX=B$, where $A$ is the $n \times n$ coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants.
• Conditions for Infinitely Many Solutions (Cramer's Rule): For a system of three linear equations in three variables ($x, y, z$) to have infinitely many solutions, the following two conditions must be satisfied:
  1. The determinant of the coefficient matrix $A$, denoted as $\det(A)$ or $\Delta$, must be zero. This indicates that the equations are linearly dependent. $\Delta = \det(A) = 0$
  2. The system must be consistent. This means that all the determinants $\Delta_x, \Delta_y, \Delta_z$ (formed by replacing the respective variable's coefficient column in $A$ with the constant vector $B$) must also be zero. $\Delta_x = 0, \quad \Delta_y = 0, \quad \Delta_z = 0$ If $\Delta=0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution (it is inconsistent).

The given system of linear equations is:

$$\begin{aligned} x+y+2z &= 6 \\ 2x+3y+az &= a+1 \\ -x-3y+bz &= 2b \end{aligned}$$

The corresponding matrix form $AX=B$ is:

$$\begin{pmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ a+1 \\ 2b \end{pmatrix}$$

Here, the coefficient matrix is $A = \begin{pmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{pmatrix}$ and the constant vector is $B = \begin{pmatrix} 6 \\ a+1 \\ 2b \end{pmatrix}$.

---

Step-by-Step Solution

Step 1: Calculate the determinant of the coefficient matrix, $\Delta$, and set it to zero. For infinitely many solutions, the first condition is $\Delta = \det(A) = 0$.

$$\Delta = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix}$$

Expanding the determinant along the first row:

$$\Delta = 1 \cdot \begin{vmatrix} 3 & a \\ -3 & b \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & a \\ -1 & b \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & 3 \\ -1 & -3 \end{vmatrix}$$ $$\Delta = 1(3b - a(-3)) - 1(2b - a(-1)) + 2(2(-3) - 3(-1))$$ $$\Delta = (3b + 3a) - (2b + a) + 2(-6 + 3)$$ $$\Delta = 3b + 3a - 2b - a + 2(-3)$$ $$\Delta = 2a + b - 6$$

Setting $\Delta = 0$: $2a + b - 6 = 0 \quad \Rightarrow \quad b = 6 - 2a \quad \text{(Equation 1)}$ This equation provides a relationship between $a$ and $b$ that must hold for infinitely many solutions (or no solution).

Step 2: Calculate $\Delta_x$ and set it to zero. The second condition for infinitely many solutions requires consistency. We start by setting $\Delta_x = 0$. $\Delta_x$ is obtained by replacing the first column of $A$ with the constant vector $B$:

$$\Delta_x = \begin{vmatrix} 6 & 1 & 2 \\ a+1 & 3 & a \\ 2b & -3 & b \end{vmatrix}$$

Expanding the determinant along the first row:

$$\Delta_x = 6 \cdot \begin{vmatrix} 3 & a \\ -3 & b \end{vmatrix} - 1 \cdot \begin{vmatrix} a+1 & a \\ 2b & b \end{vmatrix} + 2 \cdot \begin{vmatrix} a+1 & 3 \\ 2b & -3 \end{vmatrix}$$ $$\Delta_x = 6(3b - a(-3)) - 1((a+1)b - a(2b)) + 2((a+1)(-3) - 3(2b))$$ $$\Delta_x = 6(3b + 3a) - (ab + b - 2ab) + 2(-3a - 3 - 6b)$$ $$\Delta_x = 18b + 18a - (-ab + b) - 6a - 6 - 12b$$ $$\Delta_x = 18b + 18a + ab - b - 6a - 6 - 12b$$ $$\Delta_x = 12a + 5b + ab - 6$$

Setting $\Delta_x = 0$: $12a + 5b + ab - 6 = 0 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations for $a$ and $b$. We now have a system of two equations with two variables:

1. $b = 6 - 2a$
2. $12a + 5b + ab - 6 = 0$

Substitute Equation 1 into Equation 2: $12a + 5(6 - 2a) + a(6 - 2a) - 6 = 0$ $12a + 30 - 10a + 6a - 2a^2 - 6 = 0$ Combine like terms: $-2a^2 + (12a - 10a + 6a) + (30 - 6) = 0$ $-2a^2 + 8a + 24 = 0$ Divide the entire equation by $-2$ to simplify: $a^2 - 4a - 12 = 0$ Factor the quadratic equation: $(a - 6)(a + 2) = 0$ This yields two possible values for $a$: $a=6$ or $a=-2$.

Now, find the corresponding values of $b$ using Equation 1 ($b = 6 - 2a$):

• If $a=6$: $b = 6 - 2(6) = 6 - 12 = -6$. So, $(a,b) = (6, -6)$.
• If $a=-2$: $b = 6 - 2(-2) = 6 + 4 = 10$. So, $(a,b) = (-2, 10)$.

Step 4: Verify with $\Delta_y$ to confirm infinitely many solutions. It is essential to check which of these pairs actually leads to infinitely many solutions. If $\Delta=0$ and $\Delta_x=0$, but $\Delta_y \neq 0$, the system would have no solution. We need $\Delta_y = 0$ as well. Let's calculate $\Delta_y$:

$$\Delta_y = \begin{vmatrix} 1 & 6 & 2 \\ 2 & a+1 & a \\ -1 & 2b & b \end{vmatrix}$$

Expanding along the first row:

$$\Delta_y = 1((a+1)b - a(2b)) - 6(2b - a(-1)) + 2(2(2b) - (a+1)(-1))$$ $$\Delta_y = (ab + b - 2ab) - 6(2b + a) + 2(4b + a + 1)$$ $$\Delta_y = -ab + b - 12b - 6a + 8b + 2a + 2$$ $$\Delta_y = -ab - 4a - 3b + 2$$

Now, test the two pairs $(a,b)$:

• Case 1: Test $(a=6, b=-6)$ Substitute these values into $\Delta_y$: $\Delta_y = -(6)(-6) - 4(6) - 3(-6) + 2$ $\Delta_y = 36 - 24 + 18 + 2$ $\Delta_y = 32$ Since $\Delta_y = 32 \neq 0$, this pair $(a=6, b=-6)$ leads to no solution.

• Case 2: Test $(a=-2, b=10)$ Substitute these values into $\Delta_y$: $\Delta_y = -(-2)(10) - 4(-2) - 3(10) + 2$ $\Delta_y = 20 + 8 - 30 + 2$ $\Delta_y = 0$ Since $\Delta_y = 0$ (and we already have $\Delta=0$ and $\Delta_x=0$), this pair $(a=-2, b=10)$ satisfies all conditions for having infinitely many solutions. This is the correct pair of values for $a$ and $b$.

Step 5: Calculate the required expression $7a+3b$. Using the correct values $a=-2$ and $b=10$: $7a+3b = 7(-2) + 3(10)$ $7a+3b = -14 + 30$ $7a+3b = 16$

---

Common Mistakes & Tips

• Incomplete Conditions: A frequent error is to only check $\Delta=0$ and forget the consistency conditions ($\Delta_x = \Delta_y = \Delta_z = 0$). $\Delta=0$ alone indicates either no solution or infinitely many solutions, but not definitively which one.
• Algebraic Errors: Determinant calculations can be lengthy and prone to sign errors or arithmetic mistakes. Double-check each step carefully.
• Gaussian Elimination as an Alternative: While Cramer's rule is used here, an alternative approach is Gaussian elimination (row reduction). For infinitely many solutions, Gaussian elimination will lead to at least one row of zeros (e.g., $0=0$) after reducing the augmented matrix to row echelon form, provided the system is consistent. This method can sometimes be less prone to calculation errors with large determinants.

---

Summary

To find the values of $a$ and $b$ for which the system has infinitely many solutions, we applied the conditions from Cramer's Rule. First, we set the determinant of the coefficient matrix ($\Delta$) to zero, which gave us a relationship between $a$ and $b$. Next, we set $\Delta_x$ (the determinant formed by replacing the $x$-coefficients with the constant terms) to zero, which provided a second equation. Solving these two equations yielded two possible pairs for $(a,b)$. Finally, we used $\Delta_y$ (the determinant formed by replacing the $y$-coefficients with the constant terms) as a consistency check. Only the pair $(a=-2, b=10)$ resulted in $\Delta_y=0$, confirming infinitely many solutions. Substituting these values into $7a+3b$ gave the final answer.

The final answer is $\boxed{\text{16}}$, which corresponds to option (A).