$$ \begin{array}{l} ext { d all values of } a ext { for which the linear system } \ \qquad \begin{aligned} x_{1}+2 x_{2}+x_{3} & =3 \ 2 x_{1}+3 x_{2}+3 x_{3} & =4\end{aligned} \ \qquad 3 x_{1}+3 x_{2}+\left(a^{2}-3 ight) x_{3}=a\end{array} ~=~ $$ (a) no solution, (b) a unique solution and (c) infinitely many solutions.

Question

$$ \begin{array}{l}	ext { d all values of } a 	ext { for which the linear system } \ \qquad \begin{aligned} x_{1}+2 x_{2}+x_{3} & =3 \ 2 x_{1}+3 x_{2}+3 x_{3} & =4\end{aligned} \ \qquad 3 x_{1}+3 x_{2}+\left(a^{2}-3ight) x_{3}=a\end{array} ~=~ $$ (a) no solution, (b) a unique solution and (c) infinitely many solutions.

UpStudy AI · Accepted Answer

To determine the values of $$ a $$ for which the given linear system has no solution, a unique solution, or infinitely many solutions, we can analyze the system of equations:

$$
\begin{aligned}
1. & \quad x_{1} + 2x_{2} + x_{3} = 3 \quad \text{(Equation 1)} \
2. & \quad 2x_{1} + 3x_{2} + 3x_{3} = 4 \quad \text{(Equation 2)} \
3. & \quad 3x_{1} + 3x_{2} + (a^{2} - 3)x_{3} = a \quad \text{(Equation 3)}
\end{aligned}
$$

We can represent this system in matrix form as $$ A\mathbf{x} = \mathbf{b} $$, where:

$$
A = \begin{bmatrix}
1 & 2 & 1 \
2 & 3 & 3 \
3 & 3 & a^{2} - 3
\end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix}
x_{1} \
x_{2} \
x_{3}
\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}
3 \
4 \
a
\end{bmatrix}
$$

To analyze the system, we will calculate the determinant of the coefficient matrix $$ A $$ and use it to determine the conditions for the solutions.

### Step 1: Calculate the determinant of $$ A $$

The determinant of a $$ 3 \times 3 $$ matrix is given by:

$$
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
$$

For our matrix $$ A $$:

$$
\text{det}(A) = 1 \cdot (3(a^{2} - 3) - 3 \cdot 3) - 2 \cdot (2(a^{2} - 3) - 3 \cdot 3) + 1 \cdot (2 \cdot 3 - 3 \cdot 2)
$$

Calculating each term:

1. $$ 3(a^{2} - 3) - 9 = 3a^{2} - 18 $$
2. $$ 2(a^{2} - 3) - 9 = 2a^{2} - 15 $$
3. $$ 2 \cdot 3 - 3 \cdot 2 = 0 $$

Now substituting back into the determinant formula:

$$
\text{det}(A) = 1(3a^{2} - 18) - 2(2a^{2} - 15) + 0
$$

This simplifies to:

$$
\text{det}(A) = 3a^{2} - 18 - 4a^{2} + 30 = -a^{2} + 12
$$

### Step 2: Analyze the determinant

The determinant $$ \text{det}(A) = -a^{2} + 12 $$ can be set to zero to find the critical points:

$$
-a^{2} + 12 = 0 \implies a^{2} = 12 \implies a = \pm 2\sqrt{3}
$$

### Step 3: Determine the conditions for solutions

1. **No solution**: This occurs when the determinant is zero and the system is inconsistent. This happens at $$ a = \pm 2\sqrt{3} $$.
2. **Unique solution**: This occurs when the determinant is non-zero. This happens for $$ -2\sqrt{3} < a < 2\sqrt{3} $$.
3. **Infinitely many solutions**: This occurs when the determinant is zero and the system is consistent. This will not happen in this case since the equations are linearly independent.

### Conclusion

- **No solution**: $$ a = \pm 2\sqrt{3} $$
- **Unique solution**: $$ -2\sqrt{3} < a < 2\sqrt{3} $$
- **Infinitely many solutions**: There are no values of $$ a $$ for which this occurs.

Thus, the final answer is:

- (a) No solution: $$ a = \pm 2\sqrt{3} $$
- (b) A unique solution: $$ -2\sqrt{3} < a < 2\sqrt{3} $$
- (c) Infinitely many solutions: None.
For the given linear system, the values of $$ a $$ are: - **No solution**: $$ a = \pm 2\sqrt{3} $$ - **Unique solution**: $$ -2\sqrt{3} < a < 2\sqrt{3} $$ - **Infinitely many solutions**: None

(a) no solution, (b) a unique solution and © infinitely many solutions.

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