Answer :
To determine the value of [tex]\( p \)[/tex] for which the system of equations:
[tex]\[ \begin{cases} x + 2y - 3z = 1 \quad \text{(Equation 1)} \\ (p + 2)z = 3 \quad \text{(Equation 2)} \\ (2p + 1)y + z = 2 \quad \text{(Equation 3)} \end{cases} \][/tex]
has an infinite number of solutions, we need to ensure that these equations are dependent, meaning one of the equations can be expressed as a linear combination of the others.
First, let's solve Equation 2 for [tex]\( z \)[/tex]:
[tex]\[ (p + 2)z = 3 \][/tex]
So,
[tex]\[ z = \frac{3}{p + 2} \quad \text{(as long as }p \neq -2) \][/tex]
Next, substitute this value of [tex]\( z \)[/tex] into Equation 3:
[tex]\[ (2p + 1)y + \frac{3}{p + 2} = 2 \][/tex]
Isolate [tex]\( y \)[/tex]:
[tex]\[ (2p + 1)y = 2 - \frac{3}{p + 2} \][/tex]
So,
[tex]\[ y = \frac{2 - \frac{3}{p + 2}}{2p + 1} \][/tex]
Next, substitute the values of [tex]\( y \)[/tex] and [tex]\( z \)[/tex] into Equation 1 to see the consistency condition:
[tex]\[ x + 2\left(\frac{2 - \frac{3}{p + 2}}{2p + 1}\right) - 3\left(\frac{3}{p + 2}\right) = 1 \][/tex]
For the system to have an infinite number of solutions, this equation must hold for any value of [tex]\( x \)[/tex]. This means the coefficients in the system must result in dependent equations, where the combined determinant of the coefficient matrix is zero, indicating dependency of equations.
A key criteria is that [tex]\( p + 2 \neq 0 \)[/tex], since division by zero is undefined.
Thus, we find:
[tex]\[ p \neq -2 \][/tex]
This ensures that there is a consistent solution under the condition that their coefficients provide dependency required for infinite solutions.
Thus, the value of [tex]\( p \)[/tex] is not equal to:
[tex]\[ \boxed{-2} \][/tex]
[tex]\[ \begin{cases} x + 2y - 3z = 1 \quad \text{(Equation 1)} \\ (p + 2)z = 3 \quad \text{(Equation 2)} \\ (2p + 1)y + z = 2 \quad \text{(Equation 3)} \end{cases} \][/tex]
has an infinite number of solutions, we need to ensure that these equations are dependent, meaning one of the equations can be expressed as a linear combination of the others.
First, let's solve Equation 2 for [tex]\( z \)[/tex]:
[tex]\[ (p + 2)z = 3 \][/tex]
So,
[tex]\[ z = \frac{3}{p + 2} \quad \text{(as long as }p \neq -2) \][/tex]
Next, substitute this value of [tex]\( z \)[/tex] into Equation 3:
[tex]\[ (2p + 1)y + \frac{3}{p + 2} = 2 \][/tex]
Isolate [tex]\( y \)[/tex]:
[tex]\[ (2p + 1)y = 2 - \frac{3}{p + 2} \][/tex]
So,
[tex]\[ y = \frac{2 - \frac{3}{p + 2}}{2p + 1} \][/tex]
Next, substitute the values of [tex]\( y \)[/tex] and [tex]\( z \)[/tex] into Equation 1 to see the consistency condition:
[tex]\[ x + 2\left(\frac{2 - \frac{3}{p + 2}}{2p + 1}\right) - 3\left(\frac{3}{p + 2}\right) = 1 \][/tex]
For the system to have an infinite number of solutions, this equation must hold for any value of [tex]\( x \)[/tex]. This means the coefficients in the system must result in dependent equations, where the combined determinant of the coefficient matrix is zero, indicating dependency of equations.
A key criteria is that [tex]\( p + 2 \neq 0 \)[/tex], since division by zero is undefined.
Thus, we find:
[tex]\[ p \neq -2 \][/tex]
This ensures that there is a consistent solution under the condition that their coefficients provide dependency required for infinite solutions.
Thus, the value of [tex]\( p \)[/tex] is not equal to:
[tex]\[ \boxed{-2} \][/tex]