1. Key Concepts and Formulas

• System of Linear Equations: A system of linear equations can be represented in matrix form as $AX=B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants.
• Unique Solution: A system of linear equations $AX=B$ has a unique solution if and only if the determinant of the coefficient matrix $A$, denoted as $\det(A)$ or $\Delta$, is non-zero. That is, $\Delta \neq 0$.
• No Solution (Inconsistent System): A system of linear equations $AX=B$ has no solution if and only if $\Delta = 0$ AND the system is inconsistent. In terms of rank, this means $\text{rank}(A) \neq \text{rank}([A|B])$. When using row operations, this typically manifests as a row of zeros in the coefficient part of the augmented matrix corresponding to a non-zero constant on the right-hand side (e.g., $0x+0y+0z=k$ where $k \neq 0$).
• Infinitely Many Solutions (Consistent System): A system of linear equations $AX=B$ has infinitely many solutions if and only if $\Delta = 0$ AND the system is consistent. In terms of rank, this means $\text{rank}(A) = \text{rank}([A|B]) < \text{number of variables}$. When using row operations, this means a row of zeros in the coefficient part of the augmented matrix also corresponds to a zero constant on the right-hand side (e.g., $0x+0y+0z=0$).

2. Step-by-Step Solution

Step 1: Define the Sample Space Two fair dice are thrown, and the numbers on them are taken as $\lambda$ and $\mu$. This means $\lambda$ and $\mu$ can each take any integer value from 1 to 6. The total number of possible ordered pairs $(\lambda, \mu)$ is $6 \times 6 = 36$. Each outcome is equally likely.

Step 2: Construct the Coefficient Matrix and Calculate its Determinant The given system of linear equations is:

1. $x + y + z = 5$
2. $x + 2y + 3z = \mu$
3. $x + 3y + \lambda z = 1$

The coefficient matrix $A$ is: $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{pmatrix}$ Now, we calculate the determinant of $A$, denoted as $\Delta$: We can use row operations to simplify the matrix before expanding the determinant. $R_2 \to R_2 - R_1$ $R_3 \to R_3 - R_1$ $\Delta = \det \begin{pmatrix} 1 & 1 & 1 \\ 1-1 & 2-1 & 3-1 \\ 1-1 & 3-1 & \lambda-1 \end{pmatrix} = \det \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & \lambda-1 \end{pmatrix}$ Expanding along the first column (which has two zeros): $\Delta = 1 \cdot \det \begin{pmatrix} 1 & 2 \\ 2 & \lambda-1 \end{pmatrix} - 0 + 0$ $\Delta = 1 \cdot (1 \cdot (\lambda-1) - 2 \cdot 2)$ $\Delta = (\lambda-1) - 4$ $\Delta = \lambda - 5$ The determinant of the coefficient matrix is $\Delta = \lambda - 5$.

Step 3: Calculate Probability $p$ (Unique Solution) For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. So, $\Delta \neq 0$. From our calculation, $\Delta = \lambda - 5$. Thus, $\lambda - 5 \neq 0 \implies \lambda \neq 5$.

The possible values for $\lambda$ (from a die roll) are $\{1, 2, 3, 4, 5, 6\}$. The values of $\lambda$ for which $\Delta \neq 0$ are $\{1, 2, 3, 4, 6\}$. There are 5 such values. The value of $\mu$ (from the second die roll) does not affect the determinant $\Delta$, so $\mu$ can be any of its possible values: $\{1, 2, 3, 4, 5, 6\}$. There are 6 such values.

The number of favorable outcomes for a unique solution is the product of the number of favorable $\lambda$ values and the number of favorable $\mu$ values: Number of favorable outcomes = $5 \times 6 = 30$. The total number of possible outcomes is $36$.

The probability $p$ that the system has a unique solution is: $p = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{30}{36} = \frac{5}{6}$

Step 4: Calculate Probability $q$ (No Solution) For the system to have no solution, two conditions must be met:

1. $\Delta = 0$
2. The system must be inconsistent (i.e., $\text{rank}(A) \neq \text{rank}([A|B])$).

First, let's find the value(s) of $\lambda$ for which $\Delta = 0$: $\Delta = \lambda - 5 = 0 \implies \lambda = 5$. So, this condition is met for only one value of $\lambda$.

Now, we substitute $\lambda = 5$ into the augmented matrix $[A|B]$ of the system and perform row operations to check for consistency. The augmented matrix is: $[A|B] = \begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 1 & 2 & 3 & | & \mu \\ 1 & 3 & 5 & | & 1 \end{pmatrix}$ Apply row operations: $R_2 \to R_2 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & \mu-5 \\ 1 & 3 & 5 & | & 1 \end{pmatrix}$ $R_3 \to R_3 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & \mu-5 \\ 0 & 2 & 4 & | & -4 \end{pmatrix}$ $R_3 \to R_3 - 2R_2$: $\begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & \mu-5 \\ 0 & 0 & 0 & | & -4 - 2(\mu-5) \end{pmatrix}$ The last row of the augmented matrix corresponds to the equation: $0x + 0y + 0z = -4 - 2(\mu - 5)$ $0 = -4 - 2\mu + 10$ $0 = 6 - 2\mu$

• For no solution, this equation must be a contradiction, i.e., $0 = (\text{non-zero constant})$. So, $6 - 2\mu \neq 0$. $2\mu \neq 6 \implies \mu \neq 3$.

• For infinitely many solutions, this equation must be an identity, i.e., $0 = 0$. So, $6 - 2\mu = 0$. $2\mu = 6 \implies \mu = 3$.

We are interested in the probability $q$ that the system has no solution. This occurs when $\lambda = 5$ (1 value) AND $\mu \neq 3$. The possible values for $\mu$ (from a die roll) are $\{1, 2, 3, 4, 5, 6\}$. The values of $\mu$ for which $\mu \neq 3$ are $\{1, 2, 4, 5, 6\}$. There are 5 such values.

The number of favorable outcomes for no solution is: Number of favorable outcomes = (Number of $\lambda$ values for no solution) $\times$ (Number of $\mu$ values for no solution) Number of favorable outcomes = $1 \times 5 = 5$. The total number of possible outcomes is $36$.

The probability $q$ that the system has no solution is: $q = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{36}$

3. Common Mistakes & Tips

• Determinant Calculation: Be careful with signs and arithmetic when calculating the determinant. A small error here propagates through the entire problem.
• Conditions for Solutions: Remember the precise conditions: $\Delta \neq 0$ for unique, $\Delta = 0$ AND inconsistent for no solution, $\Delta = 0$ AND consistent for infinite solutions. Don't assume $\Delta=0$ automatically means no solution.
• Sample Space: Ensure you correctly identify all possible values for $\lambda$ and $\mu$ (1 to 6 for a die roll) and calculate the total number of outcomes ($6 \times 6 = 36$).
• Counting Favorable Outcomes: When $\Delta$ depends only on $\lambda$ (or $\mu$), remember that the other variable can take any of its 6 possible values when calculating the total favorable outcomes.

4. Summary

We analyzed the given system of linear equations by first calculating the determinant of its coefficient matrix, which we found to be $\Delta = \lambda - 5$. For a unique solution, $\Delta \neq 0$, which means $\lambda \neq 5$. This led to 5 favorable values for $\lambda$ and 6 for $\mu$, giving $p = 30/36 = 5/6$. For no solution, $\Delta = 0$ (meaning $\lambda = 5$) and the system must be inconsistent. By substituting $\lambda=5$ into the augmented matrix and performing row operations, we found that the system is inconsistent when $6 - 2\mu \neq 0$, which implies $\mu \neq 3$. This led to 1 favorable value for $\lambda$ and 5 for $\mu$, giving $q = 5/36$.

5. Final Answer

The calculated probabilities are $p = \frac{5}{6}$ and $q = \frac{5}{36}$. This corresponds to option (B). The final answer is $\boxed{\text{(B)}}$.