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

For a system of non-homogeneous linear equations to have infinitely many solutions, the determinant of the coefficient matrix ($\Delta$) must be zero, and the determinants $\Delta_x, \Delta_y$, and $\Delta_z$ must also be zero.
The given system is:
$x - 2y + z = 0$
$2x + 3y + z = 6$
$x + 2y + pz = q$
First, we set the determinant of the coefficient matrix to zero: $\Delta = 0$.
$\Delta = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & p \end{vmatrix} = 0$
$1(3p - 2) - (-2)(2p - 1) + 1(4 - 3) = 0$
$3p - 2 + 2(2p - 1) + 1 = 0$
$3p - 2 + 4p - 2 + 1 = 0$
$7p - 3 = 0$
$p = \frac{3}{7}$.
Next, for infinitely many solutions, $\Delta_z$ must also be zero.
$\Delta_z = \begin{vmatrix} 1 & -2 & 0 \\ 2 & 3 & 6 \\ 1 & 2 & q \end{vmatrix} = 0$
$1(3q - 12) - (-2)(2q - 6) + 0(4 - 3) = 0$
$3q - 12 + 2(2q - 6) = 0$
$3q - 12 + 4q - 12 = 0$
$7q - 24 = 0$
$q = \frac{24}{7}$.
Now we check the given options using $p = 3/7$ and $q = 24/7$.
Checking option (C): $q - p$.
$q - p = \frac{24}{7} - \frac{3}{7} = \frac{24 - 3}{7} = \frac{21}{7} = 3$.
This matches the condition given in option (C).