The system of linear equations
x + y + z = 2
2x + 3y + 2z = 5
`2x + 3y + (a^(2) - 1)z = a + 1`

Options

A. has infinitely many solutions for a = 4
B. is inconsisten when `|a| = sqrt(3)`
C. is inconsistent when a = 4
D. has a unique solution for `|a| = sqrt(3)`

Answer

To solve the system of linear equations given by:

1. \( x + y + z = 2 \) (Equation 1)
2. \( 2x + 3y + 2z = 5 \) (Equation 2)
3. \( 2x + 3y + (a^2 - 1)z = a + 1 \) (Equation 3)

we will use Cramer's Rule to determine the conditions under which the system has a unique solution, no solution, or infinitely many solutions.

### Step 1: Form the Coefficient Matrix and Calculate the Determinant \( D \)

The coefficient matrix \( A \) is given by:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 2 & 3 & a^2 - 1 \end{bmatrix} \]

To find the determinant \( D \) of matrix \( A \):

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 2 & 3 & a^2 - 1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D = 1 \cdot \begin{vmatrix} 3 & 2 \\ 3 & a^2 - 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 2 \\ 2 & a^2 - 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3 & 2 \\ 3 & a^2 - 1 \end{vmatrix} = 3(a^2 - 1) - 6 = 3a^2 - 9 \)
2. \( \begin{vmatrix} 2 & 2 \\ 2 & a^2 - 1 \end{vmatrix} = 2(a^2 - 1) - 4 = 2a^2 - 6 \)
3. \( \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 0 \)

Thus, we have:

\[ D = 1(3a^2 - 9) - 1(2a^2 - 6) + 0 \]
\[ D = 3a^2 - 9 - 2a^2 + 6 = a^2 - 3 \]

### Step 2: Determine Conditions for Solutions

For the system to have a unique solution, \( D \) must be non-zero:

\[ D \neq 0 \implies a^2 - 3 \neq 0 \implies a^2 \neq 3 \implies a \neq \pm \sqrt{3} \]

For the system to have no solution, we need to check the determinant of the augmented matrix \( D_1 \).

### Step 3: Form the Augmented Matrix and Calculate \( D_1 \)

The augmented matrix is:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 2 & 3 & 2 & | & 5 \\ 2 & 3 & a^2 - 1 & | & a + 1 \end{bmatrix} \]

The determinant \( D_1 \) is given by:

\[ D_1 = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & 5 \\ 2 & 3 & a + 1 \end{vmatrix} \]

Calculating \( D_1 \):

\[ D_1 = 1 \cdot \begin{vmatrix} 3 & 5 \\ 3 & a + 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 5 \\ 2 & a + 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3 & 5 \\ 3 & a + 1 \end{vmatrix} = 3(a + 1) - 15 = 3a - 12 \)
2. \( \begin{vmatrix} 2 & 5 \\ 2 & a + 1 \end{vmatrix} = 2(a + 1) - 10 = 2a - 8 \)
3. \( \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 0 \)

Thus, we have:

\[ D_1 = 1(3a - 12) - 1(2a - 8) + 0 \]
\[ D_1 = 3a - 12 - 2a + 8 = a - 4 \]

### Step 4: Conditions for No Solution

For the system to have no solution, we need:

\[ D = 0 \quad \text{and} \quad D_1 \neq 0 \]

From \( D = 0 \):

\[ a^2 - 3 = 0 \implies a = \pm \sqrt{3} \]

From \( D_1 \neq 0 \):

\[ a - 4 \neq 0 \implies a \neq 4 \]

### Conclusion

Thus, the system has:

- A unique solution when \( a \neq \pm \sqrt{3} \) and \( a \neq 4 \).
- No solution when \( a = \sqrt{3} \) or \( a = -\sqrt{3} \) and \( a \neq 4 \).