Question

Consider the following system of linear equations: 
x + 2y + 3z = 0 
2x + py = 0 
3x + 2y + pz = 0
The value(s) of \( p \) for which the system of equations have infinitely many solutions is/are:

• A. \( p = 1 \)
• B. \( p = 2 \)
• C. \( p = 6 \)
• D. \( p = 12 \)

Hint:

For systems of linear equations, when the determinant of the coefficient matrix is zero, the system may have infinitely many solutions, depending on the consistency of the equations. Check for values of \( p \) that make the determinant zero to find such cases.

Answer 1

The Correct Option is A, D

We are given the following system of linear equations: x + 2y + 3z = 0 
2x + py = 0 
3x + 2y + pz = 0
Represent the system in augmented matrix form:

\[ \begin{bmatrix} 1 & 2 & 3 & | & 0 \\ 2 & p & 0 & | & 0 \\ 3 & 2 & p & | & 0 \end{bmatrix} \]

Perform row operations to simplify the matrix. 
Step 1: Subtract \( 2 \times \) Row 1 from Row 2:

\[ \begin{bmatrix} 1 & 2 & 3 & | & 0 \\ 0 & p-4 & -6 & | & 0 \\ 3 & 2 & p & | & 0 \end{bmatrix} \]

Next, subtract \( 3 \times \) Row 1 from Row 3:

\[ \begin{bmatrix} 1 & 2 & 3 & | & 0 \\ 0 & p-4 & -6 & | & 0 \\ 0 & -4 & p-9 & | & 0 \end{bmatrix} \]

Step 2: To find when the system has infinite solutions, we calculate the determinant of the coefficient matrix:

\[ \begin{vmatrix} 1 & 2 & 3 \\ 0 & p-4 & -6 \\ 0 & -4 & p-9 \end{vmatrix} = 1 \cdot \begin{vmatrix} p-4 & -6 \\ -4 & p-9 \end{vmatrix} \]

Expanding the determinant:

\[ = (p-4)(p-9) - (-6)(-4) \]

\[ = p^2 - 13p + 36 - 24 \]

\[ = p^2 - 13p + 12 \]

Set the determinant equal to 0:

\[ p^2 - 13p + 12 = 0 \]

Factoring:

\[ (p - 12)(p - 1) = 0 \]

Thus, \( p = 12 \) or \( p = 1 \). 
Step 3: Check the consistency of the system for these values of \( p \). Both \( p = 1 \) and \( p = 12 \) make the system dependent, meaning there are infinitely many solutions. 
Therefore, the correct answer is: \( p = 1 \) and \( D \) \( p = 12 \)