Question

If the system of simultaneous linear equations $x - 2y + z = 0$, $2x + 3y + z = 6$ and $x + 2y + pz = q$ has infinitely many solutions, then

• A. $p + q = 4$
• B. $pq = 48/49$
• C. $q - p = 3$
• D. $p/q = 4/9$

Hint:

A system of linear equations $AX = B$ has infinitely many solutions if and only if the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A|B]$, and this rank is less than the number of variables. For a 3x3 system, this is equivalent to $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$.

Answer 1

The Correct Option is C

To determine the conditions under which the system of equations has infinitely many solutions, we consider the given set of linear equations:

• \(x - 2y + z = 0\)
• \(2x + 3y + z = 6\)
• \(x + 2y + pz = q\)

For a system of linear equations to have infinitely many solutions, the equations need to be linearly dependent. This implies that each equation is a linear combination of the others.

First, let's express these equations in matrix form:

\(\begin{bmatrix} 1 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & p \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ q \end{bmatrix}\)

The coefficient matrix is:

\(A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & p \end{bmatrix}\)

To ensure the system has infinitely many solutions, the rank of the augmented matrix \([A|b]\) must be equal to the rank of the coefficient matrix \(A\), but less than the number of variables, i.e., rank should be 2 in this 3-variable system for non-trivial solutions.

The augmented matrix is:

\([A|b] = \begin{bmatrix} 1 & -2 & 1 & 0 \\ 2 & 3 & 1 & 6 \\ 1 & 2 & p & q \end{bmatrix}\)

We perform row operations on the augmented matrix to check for consistency:

• Subtract the first row from the third: \([0, 4, p-1, q]\)
• Subtract twice the first row from the second: \([0, 7, -1, 6]\)
• New matrix:

\(\begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 7 & -1 & 6 \\ 0 & 4 & p-1 & q \end{bmatrix}\)

For the matrix to have rank 2 (indicating infinitely many solutions), the third row must become a linear combination of the second, leading to no new information. Therefore:

\(k[0, 7, -1, 6] = [0, 4, p-1, q]\)

Equating coefficients:

• \(7k = 4 \Rightarrow k = \frac{4}{7}\)
• \(-k = p - 1 \Rightarrow -\frac{4}{7} = p - 1 \Rightarrow p = \frac{3}{7}\)
• \(6k = q \Rightarrow q = \frac{24}{7}\)

Now solving \(q - p\):

\(q - p = \frac{24}{7} - \frac{3}{7} = \frac{21}{7} = 3\)

Thus, the correct option is indeed \(q - p = 3\).