For some constant real numbers p,k and a consider the following system of linear equations in x and y:
px−4y=23x+ky=a
A necessary condition for the system to have no solution for (x,y) is
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Official Correct Answer: C. The system of equations given is:
1. \( p x - 4 y = 2 \)
2. \( 3 x + k y = a \)
And the question is asking for a necessary condition such that this system has no solution. The options are A, B, C, D, with C being the correct answer according to the official solution. But I need to figure out why.
First, for a system of linear equations to have no solution, the two lines must be parallel but not coinciding. In other words, their slopes must be equal, but their y-intercepts must be different. Alternatively, in terms of linear algebra, the determinant of the coefficient matrix should be zero, and the determinant of the augmented matrix should not be zero.
We have the system again:
\( p x - 4 y = 2 \) ...(1)
\( 3 x + k y = a \) ...(2)
To analyze this, I can represent this system in matrix form:
\[
\begin{bmatrix}
p & -4 \\
3 & k \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
\end{bmatrix}
=
\begin{bmatrix}
2 \\
a \\
\end{bmatrix}
\]
For a system \( A \mathbf{x} = \mathbf{b} \), the system has no solution if the rank of the coefficient matrix \( A \) is less than the rank of the augmented matrix \( [A|\mathbf{b}] \). This happens when the determinant of \( A \) is zero, and the determinant of the augmented matrix is non-zero.
So, first, Computing the determinant of the coefficient matrix \( A \):
\[
\text{det}(A) = (p)(k) - (-4)(3) = pk + 12
\]
For the system to have no solution, the determinant must be zero. So,
\( pk + 12 = 0 \) ...(3)
But that's just the first condition. The second condition is that the augmented matrix must have a higher rank, which means the equations are inconsistent.
To check for inconsistency, we can use the concept of cross-multiplication or ratios. If the ratios of the coefficients are equal, but the ratio of the constants is different, then the system is inconsistent.
So, let's set up the ratios:
The coefficients of x: \( p \) and \( 3 \)
The coefficients of y: \( -4 \) and \( k \)
The constants: \( 2 \) and \( a \)
For the system to be inconsistent, the following must hold:
\( \frac{p}{3} = \frac{-4}{k} \neq \frac{2}{a} \)
From the first part, \( \frac{p}{3} = \frac{-4}{k} \), which implies \( pk = -12 \), which is consistent with equation (3) since \( pk + 12 = 0 \) leads to \( pk = -12 \).
Now, the second part is \( \frac{p}{3} \neq \frac{2}{a} \). Let's express this inequality.
From \( \frac{p}{3} \neq \frac{2}{a} \), cross-multiplying gives \( p a \neq 6 \).
So, the necessary condition is \( pa \neq 6 \). But looking at the options, I don't see an option that directly states \( pa \neq 6 \).
Options:
A) \( a p - 6 = 0 \)
B) 2
C) \( 2a + k \neq 0 \)
D) \( k p + 12 \neq 0 \)
Hmm, none of these directly say \( pa \neq 6 \). But let's think differently. Maybe I can express \( pa \neq 6 \) in terms of other variables.
From equation (3), we have \( pk = -12 \). So, \( k = -12 / p \).
Substituting \( k \) into the inequality \( pa \neq 6 \).
But wait, maybe I should approach this differently. Let's consider the condition for inconsistency using determinants.
Another method is to use Cramer's Rule. For the system to have no solution, the determinant of the coefficient matrix must be zero, and at least one of the determinants for x or y must be non-zero.
But perhaps a more straightforward approach is to use the concept of parallel lines.
The two equations can be rewritten in slope-intercept form to find their slopes.
From equation (1):
\( p x - 4 y = 2 \)
=> \( -4 y = -p x + 2 \)
=> \( y = (p/4) x - (2/4) \)
=> \( y = (p/4) x - 1/2 \)
So, the slope \( m_1 = p/4 \)
From equation (2):
\( 3 x + k y = a \)
=> \( k y = -3 x + a \)
=> \( y = (-3/k) x + a/k \)
So, the slope \( m_2 = -3/k \)
For the lines to be parallel, their slopes must be equal:
\( p/4 = -3/k \)
Cross-multiplying:
\( p k = -12 \)
Which is the same as equation (3). So, that's consistent.
Now, for the lines to be parallel but not coinciding, their y-intercepts must be different.
From equation (1), y-intercept is \( -1/2 \)
From equation (2), y-intercept is \( a/k \)
So, for the lines to not coincide, \( -1/2 \neq a/k \)
Which implies \( a/k \neq -1/2 \)
Multiplying both sides by k (assuming k ≠ 0, which it must be because if k were 0, the second equation would be 3x = a, which is a vertical line, and the first equation is a line with slope p/4. Unless p is also 0, which would make the first equation -4y=2, which is a horizontal line. But in that case, if p=0 and k=0, the system would either have no solution or infinitely many, depending on a. But since we're looking for no solution, let's assume k ≠ 0 for now.)
So, \( a/k \neq -1/2 \) => \( a ≠ -k/2 \)
But how does this relate to the options given?
Looking at the options, option C is \( 2a + k ≠ 0 \).