Consider a system of linear equations $3x + y - z = 0$, $x - \dfrac{py}{4} + z = 2$ and $2x - y + 2z = q$ where $p, q \in I$ and $p, q \in [1, 10]$, then identify the correct statement(s).
Step-by-Step Solution
Key Concept: The determinant of the coefficient matrix determines uniqueness. When the determinant is zero, the augmented matrix must be analyzed for consistency (no solution vs. infinitely many solutions).
Step 1:
To analyze the given system of linear equations, we first write down the equations and identify the coefficient matrix. The system of equations is given by $3x + y - z = 0$, $x - \dfrac{py}{4} + z = 2$, and $2x - y + 2z = q$, where $p, q \in I$ and $p, q \in [1, 10]$. The coefficient matrix $A$ for this system is:
$$A = \begin{pmatrix} 3 & 1 & -1 \\ 1 & -p/4 & 1 \\ 2 & -1 & 2 \end{pmatrix}$$
Step 2:
Next, we need to compute the determinant of the coefficient matrix $A$, denoted as $\det(A)$. This is done by expanding the determinant along the first row. The formula for $\det(A)$ is:
$$\det(A) = 3\left[(-p/4)(2) - (1)(-1)\right] - 1\left[(1)(2) - (1)(2)\right] + (-1)\left[(1)(-1) - (-p/4)(2)\right]$$
Simplifying this expression gives us the determinant in terms of $p$.
Step 3:
Simplifying the determinant expression:
$$\det(A) = 3\left[-p/2 + 1\right] - 1\left[2 - 2\right] + (-1)\left[-1 + p/2\right]$$
$$= 3(1 - p/2) - 0 - (-1 + p/2)$$
$$= 3 - 3p/2 + 1 - p/2$$
$$= 4 - 2p$$
So, the determinant of $A$ is $\det(A) = 4 - 2p$.
Step 4:
Now, we analyze the condition for the system to have a unique solution. A unique solution exists if and only if $\det(A) \neq 0$. This implies $4 - 2p \neq 0$, which gives us $p \neq 2$. Since $p \in \{1,2,\ldots,10\}$, there are 9 values of $p$ for which $\det(A) \neq 0$. For each such $p$, $q$ can be any of 10 values. Therefore, the number of ordered pairs $(p, q)$ that result in a unique solution is $9 \times 10 = 90$.
Step 5:
We then consider the singular case where $\det(A) = 0$, which implies $p = 2$. Substituting $p = 2$ into the original system of equations gives us:
$$3x + y - z = 0$$
$$x - \frac{y}{2} + z = 2$$
$$2x - y + 2z = q$$
We notice that equation (3) is equivalent to $2 \times$ equation (2), which means $2x - y + 2z = 4$ for consistency. Thus, for the system to be consistent when $p = 2$, we must have $q = 4$.
Step 6:
To further analyze the case when $p = 2$ and $q = 4$, we examine the augmented matrix:
$$\begin{pmatrix} 3 & 1 & -1 & 0 \\ 1 & -1/2 & 1 & 2 \\ 2 & -1 & 2 & 4 \end{pmatrix}$$
Performing row operations, specifically $R_3 \to R_3 - 2R_2$, results in:
$$(0, 0, 0, 4-4) = (0, 0, 0, 0)$$
This confirms that the system is consistent when $q = 4$. Additionally, row reducing $R_1$ and $R_2$ yields a non-trivial result, indicating that the system has a free variable and thus infinitely many solutions when $q = 4$.
Step 7:
Considering all cases, we find that there are 90 ordered pairs $(p, q)$ that result in a unique solution, 1 ordered pair that results in infinitely many solutions (when $p = 2$ and $q = 4$), and 9 ordered pairs that result in no solution (when $p = 2$ and $q \neq 4$). The statement regarding at least one solution includes both unique and infinite solutions, totaling 91 ordered pairs.
Step 8:
Finally, we conclude that the correct statements regarding the system of linear equations are based on the number of ordered pairs $(p, q)$ that satisfy the conditions for unique solutions, infinite solutions, no solutions, and at least one solution. Given the calculations, the correct answer is related to the number of ordered pairs for each scenario. The final answer is: $\boxed{1}$
Correct Answer: 1