Let the system of linear equations
`x+y+kz=2`
`2x+3y-z=1`
`3x+4y+2z=k`
have infinitely many solutions.Then the system
`(k+1)x+(2k-1)y=7`
`(2k+1)x+(k+5y)=10` has :

Options

A. infinitely many solution
B. unique solution satisfying x – y = 1
C. no solution
D. unique solution satisfying x + y = 1

Answer

To solve the problem step by step, we need to analyze the given system of linear equations and determine the conditions under which they have infinitely many solutions.

### Step 1: Write the system of equations
We have the following system of equations:
1. \( x + y + kz = 2 \) (Equation 1)
2. \( 2x + 3y - z = 1 \) (Equation 2)
3. \( 3x + 4y + 2z = k \) (Equation 3)

### Step 2: Formulate the augmented matrix
The augmented matrix for the system can be formed as follows:
\[ \begin{bmatrix} 1 & 1 & k & | & 2 \\ 2 & 3 & -1 & | & 1 \\ 3 & 4 & 2 & | & k \end{bmatrix} \]

### Step 3: Calculate the determinant of the coefficient matrix
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix is:
\[ \begin{bmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{bmatrix} \]
We calculate the determinant:
\[ \text{det} = 1 \cdot (3 \cdot 2 - (-1) \cdot 4) - 1 \cdot (2 \cdot 2 - (-1) \cdot 3) + k \cdot (2 \cdot 4 - 3 \cdot 3) \]
Calculating each term:
1. \( 1 \cdot (6 + 4) = 10 \)
2. \( -1 \cdot (4 + 3) = -7 \)
3. \( k \cdot (8 - 9) = -k \)

Putting it all together:
\[ \text{det} = 10 - 7 - k = 3 - k \]
Setting the determinant to zero for infinitely many solutions:
\[ 3 - k = 0 \implies k = 3 \]

### Step 4: Substitute \( k \) back into the equations
Now substitute \( k = 3 \) into the original equations:
1. \( x + y + 3z = 2 \)
2. \( 2x + 3y - z = 1 \)
3. \( 3x + 4y + 6z = 3 \)

### Step 5: Analyze the second system of equations
Now we analyze the second system:
1. \( (k+1)x + (2k-1)y = 7 \)
2. \( (2k+1)x + (k+5)y = 10 \)

Substituting \( k = 3 \):
1. \( (3+1)x + (2 \cdot 3 - 1)y = 7 \) simplifies to \( 4x + 5y = 7 \)
2. \( (2 \cdot 3 + 1)x + (3 + 5)y = 10 \) simplifies to \( 7x + 8y = 10 \)

### Step 6: Solve the new system of equations
We now have the system:
1. \( 4x + 5y = 7 \) (Equation 4)
2. \( 7x + 8y = 10 \) (Equation 5)

To solve these equations, we can use the elimination method. Multiply Equation 4 by 7 and Equation 5 by 4:
1. \( 28x + 35y = 49 \) (Equation 6)
2. \( 28x + 32y = 40 \) (Equation 7)

Now subtract Equation 7 from Equation 6:
\[ (28x + 35y) - (28x + 32y) = 49 - 40 \]
This simplifies to:
\[ 3y = 9 \implies y = 3 \]

Substituting \( y = 3 \) back into Equation 4:
\[ 4x + 5(3) = 7 \implies 4x + 15 = 7 \implies 4x = -8 \implies x = -2 \]

### Step 7: Conclusion
The solution to the second system of equations is \( x = -2 \) and \( y = 3 \). Since we have found a unique solution, we conclude that the second system has a unique solution.

### Final Answer
The second system of equations has a unique solution.