$$ \begin{array}{l} ext { d all values of } a ext { for which the linear system } \ \qquad \begin{aligned} x_{1}+2 x_{2}+x_{3} & =3 \ 2 x_{1}+3 x_{2}+3 x_{3} & =4\end{aligned} \ \qquad 3 x_{1}+3 x_{2}+\left(a^{2}-3 ight) x_{3}=a\end{array} ~=~ $$ (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

We start with the linear system

$$
\begin{array}{rcl}
x_1 + 2x_2 + x_3 & = & 3, \
2x_1 + 3x_2 + 3x_3 & = & 4, \
3x_1 + 3x_2 + (a^2-3)x_3 & = & a.
\end{array}
$$

**Step 1. Solve the first two equations.**

From the first equation, express $$x_1$$ in terms of $$x_2$$ and $$x_3$$:

$$
x_1 = 3 - 2x_2 - x_3.
$$

Substitute $$x_1$$ into the second equation:

$$
2(3-2x_2-x_3)+3x_2+3x_3=4.
$$

Simplify the equation:

$$
6 - 4x_2 - 2x_3 + 3x_2 + 3x_3 = 4,
$$

$$
6 - x_2 + x_3 = 4.
$$

Solve for $$x_2$$:

$$
-x_2 = 4 - 6 - x_3 = -2 - x_3,
$$

$$
x_2 = 2 + x_3.
$$

Now substitute $$x_2 = 2+x_3$$ back into

$$
x_1 = 3 - 2x_2 - x_3:
$$

$$
x_1 = 3 - 2(2+x_3) - x_3 = 3 - 4 - 2x_3 - x_3 = -1 - 3x_3.
$$

Thus, the general solution of the first two equations is given by

$$
x_1 = -1 - 3t,\quad x_2 = 2 + t,\quad x_3 = t,
$$

where $$t$$ is a free parameter.

**Step 2. Substitute into the third equation.**

Substitute $$x_1 = -1-3t$$, $$x_2 = 2+t$$ and $$x_3 = t$$ into the third equation:

$$
3(-1-3t) + 3(2+t) + (a^2-3)t = a.
$$

Simplify:

$$
-3 - 9t + 6 + 3t + (a^2-3)t = a.
$$

Combine like terms:

$$
3 + \bigl[-9t+3t+(a^2-3)t\bigr] = a.
$$

Calculate the coefficient of $$t$$:

$$
-9t+3t=( -6t), \quad \text{so} \quad -6t + (a^2-3)t = (a^2-9)t.
$$

Thus, we have

$$
3 + (a^2-9)t = a.
$$

Solve for $$t$$:

$$
(a^2-9)t = a - 3.
$$

Now, we consider two main cases.

**Case (i): $$a^2-9 \neq 0$$.**

Since

$$
a^2-9 = (a-3)(a+3),
$$

this condition means $$a\neq 3$$ and $$a\neq -3$$. Then we have

$$
t = \frac{a-3}{a^2-9} = \frac{a-3}{(a-3)(a+3)} = \frac{1}{a+3} \quad \text{(provided $$a\neq 3$$)}.
$$

In this case, $$t$$ (and hence $$x_1, x_2, x_3$$) is uniquely determined. Therefore, the system has a unique solution.

**Case (ii): $$a^2-9 = 0$$.**

This gives $$a=3$$ or $$a=-3$$.

- **Subcase (a): $$a=3$$.**

Substitute $$a=3$$ into

$$
  3 + (a^2-9)t = a:
  $$
  
  Since $$3^2-9 = 9-9=0$$, the equation becomes
  
  $$
  3 = 3.
  $$
  
  This identity holds for all $$t$$. Hence, there is no new restriction on $$t$$ and infinitely many solutions occur.

- **Subcase (b): $$a=-3$$.**

For $$a=-3$$, note that $$(-3)^2-9=9-9=0$$ and the equation becomes
  
  $$
  3 = -3,
  $$
  
  which is a contradiction. Therefore, the system has no solution for $$a=-3$$.

**Answer Summary:**

- (a) The system has no solution when $$a = -3$$.
- (b) The system has a unique solution when $$a \neq 3$$ and $$a \neq -3$$.
- (c) The system has infinitely many solutions when $$a = 3$$.
The system has: - (a) no solution when $$ a = -3 $$, - (b) a unique solution when $$ a \neq 3 $$ and $$ a \neq -3 $$, - (c) infinitely many solutions when $$ a = 3 $$.

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