For what values of p and q the system of equations
`2x+5y + pz=q`
` x+2y + 3z = 14`
`x+y+z=6` is consistent ?

Answer

To determine the values of \( p \) and \( q \) for which the system of equations is consistent, we need to analyze the given equations:

1. \( 2x + 5y + pz = q \) (Equation 1)
2. \( x + 2y + 3z = 14 \) (Equation 2)
3. \( x + y + z = 6 \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Calculate the Determinant

The coefficient matrix for the system can be represented as:

\[ A = \begin{bmatrix} 2 & 5 & p \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix} \]

To find the values of \( p \) and \( q \) for which the system is consistent, we first need to calculate the determinant \( \Delta \) of the coefficient matrix \( A \).

\[ \Delta = \begin{vmatrix} 2 & 5 & p \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 2: Expand the Determinant

We can expand the determinant along the first row:

\[ \Delta = 2 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - 5 \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} + p \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = (2 \cdot 1) - (3 \cdot 1) = 2 - 3 = -1 \)

2. \( \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = (1 \cdot 1) - (3 \cdot 1) = 1 - 3 = -2 \)

3. \( \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1 \cdot 1) - (2 \cdot 1) = 1 - 2 = -1 \)

Substituting these values back into the determinant:

\[ \Delta = 2(-1) - 5(-2) + p(-1) \]
\[ \Delta = -2 + 10 - p \]
\[ \Delta = 8 - p \]

### Step 3: Conditions for Consistency

For the system to be consistent, we have two cases:

1. **Unique Solution**: This occurs when \( \Delta \neq 0 \).
\[ 8 - p \neq 0 \implies p \neq 8 \]

2. **Infinitely Many Solutions**: This occurs when \( \Delta = 0 \).
\[ 8 - p = 0 \implies p = 8 \]

### Step 4: Finding Corresponding \( q \) Values

#### Case 1: \( p \neq 8 \)

If \( p \neq 8 \), the system has a unique solution for any value of \( q \). Thus, \( q \) can be any real number.

#### Case 2: \( p = 8 \)

If \( p = 8 \), we need to ensure that the system has infinitely many solutions. We will set up the augmented matrix and find the conditions on \( q \):

\[ \begin{bmatrix} 2 & 5 & 8 & | & q \\ 1 & 2 & 3 & | & 14 \\ 1 & 1 & 1 & | & 6 \end{bmatrix} \]

We need to find the determinant of the modified matrix (let's call it \( \Delta_1 \)):

\[ \Delta_1 = \begin{vmatrix} 2 & 5 & 8 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_1 = 2 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - 5 \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} + 8 \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Using the previously calculated determinants:

\[ \Delta_1 = 2(-1) - 5(-2) + 8(-1) \]
\[ \Delta_1 = -2 + 10 - 8 = 0 \]

Now, we need to ensure that the corresponding determinant of the constants (let's call it \( \Delta_2 \)) is also zero for infinitely many solutions:

\[ \Delta_2 = \begin{vmatrix} q & 5 & 8 \\ 14 & 2 & 3 \\ 6 & 1 & 1 \end{vmatrix} \]

Calculating this determinant and setting it to zero will give us the condition for \( q \):

After performing the calculations, we find:

\[ q = 36 \]

### Final Values

Thus, the values for \( p \) and \( q \) for which the system is consistent are:

1. For \( p \neq 8 \): \( q \) can be any real number.
2. For \( p = 8 \): \( q = 36 \).