1. Key Concepts and Formulas

For a system of three linear equations in three variables $x, y, z$, represented in matrix form as $A\mathbf{x} = \mathbf{b}$:

• $A$ is the $3 \times 3$ coefficient matrix.
• $\mathbf{x}$ is the column vector of variables $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$.
• $\mathbf{b}$ is the column vector of constants.

To analyze the nature of solutions (unique, no solution, infinitely many solutions), we use determinants:

• $D = \det(A)$ is the determinant of the coefficient matrix.
• $D_1$ is the determinant formed by replacing the first column of $A$ with $\mathbf{b}$.
• $D_2$ is the determinant formed by replacing the second column of $A$ with $\mathbf{b}$.
• $D_3$ is the determinant formed by replacing the third column of $A$ with $\mathbf{b}$.

The condition for the system to have infinitely many solutions is: $D = 0 \quad \text{AND} \quad D_1 = 0 \quad \text{AND} \quad D_2 = 0 \quad \text{AND} \quad D_3 = 0$

2. Step-by-Step Solution

We are given the system of linear equations: $x+y+a z=b \quad \ldots(1)$ $2 x+5 y+2 z=6 \quad \ldots(2)$ $x+2 y+3 z=3 \quad \ldots(3)$ This system is stated to have infinitely many solutions. Our goal is to find $a$ and $b$ using the conditions mentioned above, and then calculate $2a+3b$.

Step 1: Formulate the Coefficient Matrix and Constant Vector

First, we extract the coefficient matrix $A$ and the constant vector $\mathbf{b}$ from the given system: $A = \begin{pmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} b \\ 6 \\ 3 \end{pmatrix}$

Now, we define the determinants required for Cramer's Rule: $D = \begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix}$ $D_1 = \begin{vmatrix} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix}$ $D_2 = \begin{vmatrix} 1 & b & a \\ 2 & 6 & 2 \\ 1 & 3 & 3 \end{vmatrix}$ $D_3 = \begin{vmatrix} 1 & 1 & b \\ 2 & 5 & 6 \\ 1 & 2 & 3 \end{vmatrix}$

Step 2: Apply $D=0$ to find 'a'

For infinitely many solutions, the first condition is $D=0$. We calculate the determinant $D$ by expanding along the first row: $D = 1 \begin{vmatrix} 5 & 2 \\ 2 & 3 \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ 1 & 3 \end{vmatrix} + a \begin{vmatrix} 2 & 5 \\ 1 & 2 \end{vmatrix}$ $D = 1((5)(3) - (2)(2)) - 1((2)(3) - (2)(1)) + a((2)(2) - (5)(1))$ $D = 1(15 - 4) - 1(6 - 2) + a(4 - 5)$ $D = 1(11) - 1(4) + a(-1)$ $D = 11 - 4 - a$ $D = 7 - a$

Now, we set $D=0$ to find the value of $a$: $7 - a = 0 \implies a = 7$ So, the value of $a$ is $7$.

Step 3: Apply $D_3=0$ to find 'b'

With $a=7$, we need to find $b$. We can use any of $D_1=0$, $D_2=0$, or $D_3=0$. Let's choose $D_3=0$ as it involves $b$ in the last column and the other entries are numerical, potentially simplifying calculations. We calculate $D_3$ by expanding along the first row: $D_3 = \begin{vmatrix} 1 & 1 & b \\ 2 & 5 & 6 \\ 1 & 2 & 3 \end{vmatrix}$ $D_3 = 1 \begin{vmatrix} 5 & 6 \\ 2 & 3 \end{vmatrix} - 1 \begin{vmatrix} 2 & 6 \\ 1 & 3 \end{vmatrix} + b \begin{vmatrix} 2 & 5 \\ 1 & 2 \end{vmatrix}$ $D_3 = 1((5)(3) - (6)(2)) - 1((2)(3) - (6)(1)) + b((2)(2) - (5)(1))$ $D_3 = 1(15 - 12) - 1(6 - 6) + b(4 - 5)$ $D_3 = 1(3) - 1(0) + b(-1)$ $D_3 = 3 - b$

Now, we set $D_3=0$ to find the value of $b$: $3 - b = 0 \implies b = 3$ So, the value of $b$ is $3$.

Step 4: Verify $D_1=0$ and $D_2=0$ (Crucial for Confirmation)

Although we have found $a$ and $b$ using $D=0$ and $D_3=0$, for the system to have infinitely many solutions, all four determinants must be zero. It's a good practice to verify $D_1=0$ and $D_2=0$ with $a=7$ and $b=3$.

Let's verify $D_1=0$: Substitute $a=7$ and $b=3$ into $D_1$: $D_1 = \begin{vmatrix} 3 & 1 & 7 \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix}$ $D_1 = 3 \begin{vmatrix} 5 & 2 \\ 2 & 3 \end{vmatrix} - 1 \begin{vmatrix} 6 & 2 \\ 3 & 3 \end{vmatrix} + 7 \begin{vmatrix} 6 & 5 \\ 3 & 2 \end{vmatrix}$ $D_1 = 3(15 - 4) - 1(18 - 6) + 7(12 - 15)$ $D_1 = 3(11) - 1(12) + 7(-3)$ $D_1 = 33 - 12 - 21$ $D_1 = 33 - 33 = 0$ The condition $D_1=0$ is satisfied.

Let's verify $D_2=0$: Substitute $a=7$ and $b=3$ into $D_2$: $D_2 = \begin{vmatrix} 1 & 3 & 7 \\ 2 & 6 & 2 \\ 1 & 3 & 3 \end{vmatrix}$ Observe that the second column ($C_2$) is $3 \times C_1$. When two columns (or rows) of a determinant are proportional, the determinant is zero. $D_2 = 1 \begin{vmatrix} 6 & 2 \\ 3 & 3 \end{vmatrix} - 3 \begin{vmatrix} 2 & 2 \\ 1 & 3 \end{vmatrix} + 7 \begin{vmatrix} 2 & 6 \\ 1 & 3 \end{vmatrix}$ $D_2 = 1(18 - 6) - 3(6 - 2) + 7(6 - 6)$ $D_2 = 1(12) - 3(4) + 7(0)$ $D_2 = 12 - 12 + 0 = 0$ The condition $D_2=0$ is also satisfied. Since all four determinants are zero, our values $a=7$ and $b=3$ are correct.

Step 5: Calculate the final expression

The problem asks for the value of $2a+3b$. Substitute the values $a=7$ and $b=3$ into the expression: $2a + 3b = 2(7) + 3(3)$ $2a + 3b = 14 + 9$ $2a + 3b = 23$

3. Common Mistakes & Tips

• Incomplete Conditions: A common mistake is only setting $D=0$. Remember that for infinitely many solutions, all four determinants ($D, D_1, D_2, D_3$) must be zero. If $D=0$ but any $D_j \neq 0$, the system has no solution.
• Arithmetic Errors: Determinant calculations can be prone to sign errors or simple arithmetic mistakes. Double-check your expansions.
• Strategic Choice of $D_j$: When finding the second unknown, choose the $D_j$ that appears simplest to calculate (e.g., $D_3$ in this case, as $b$ was in the last column).

4. Summary

We were given a system of linear equations stated to have infinitely many solutions. We applied the condition that for such a system, all four determinants ($D, D_1, D_2, D_3$) must be zero. First, we calculated $D$ and set it to zero, which gave us $a=7$. Next, we calculated $D_3$ and set it to zero, which yielded $b=3$. We then verified that these values of $a$ and $b$ also made $D_1$ and $D_2$ equal to zero, confirming our findings. Finally, we substituted $a=7$ and $b=3$ into the expression $2a+3b$ to get the result.

The final answer is $\boxed{23}$, which corresponds to option (A).