Question

4. a) The Gauss elimination method is used to solve the system of equations
6 x 4x x 1 + 2 + α 3 =
3 2x x 2 x 1 − 2 + α 3 =
5 x 3x x α 1 + 2 + 3 =
Find the value of α for which the system has (i) a unique solution (ii) no solution (iii)
infinitely many solutions.

Answer 1

Answer on Question #60687 – Math – Linear Algebra

Question

4. a) The Gauss elimination method is used to solve the system of equations

$$\begin{array}{l} x _ {1} + 4 x _ {2} + \alpha x _ {3} = 6 \\ 2 x _ {1} - x _ {2} + 2 \alpha x _ {3} = 3 \\ \alpha x _ {1} + 3 x _ {2} + x _ {3} = 5 \\ \end{array}$$

Find the value of $\alpha$ for which the system has

(i) a unique solution

(ii) no solution

(iii) infinitely many solutions.

Solution

Using the Gauss elimination method

\begin{array}{l} \left( \begin{array}{cccc} 1 & 4 & \alpha & 6 \\ 2 & -1 & 2\alpha & 3 \\ \alpha & 3 & 1 & 5 \end{array} \right) \to |R_2 \leftarrow R_2 - 2R_1| \to \left( \begin{array}{cccc} 1 & 4 & \alpha & 6 \\ 0 & -9 & 0 & -9 \\ \alpha & 3 & 1 & 5 \end{array} \right) \to |R_2 \leftarrow R_2 / (-9)| \to \\ \rightarrow \left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ \alpha & 3 & 1\end{array}\right) \rightarrow |R_3 \leftarrow R_3 - \alpha R_1| \rightarrow \left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 3 - 4\alpha & 1 - \alpha^2\end{array}\right) \begin{array}{c}6\\ 1\\ 5 - 6\alpha\end{array}\right) \rightarrow \\ \rightarrow | R _ {3} \leftarrow R _ {3} - (3 - 4 \alpha) R _ {2} | \rightarrow \left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & 1 - \alpha^ {2}\end{array}\right) \rightarrow \\ \rightarrow \left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & 1 - \alpha^ {2}\end{array}\right) \Bigg | _ {2 - 2 \alpha} \rightarrow \left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & (1 - \alpha) (1 + \alpha)\end{array}\right) \Bigg | _ {2 (1 - \alpha)}. \\ \end{array}

Case 1. Let $\alpha = 1$ . Then

$$\left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & (1 - \alpha) (1 + \alpha)\end{array}\right|_{2(1 - \alpha)}^{6}\rightarrow\left(\begin{array}{ccc}1 & 4 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0\end{array}\right)\rightarrow|R_1\leftarrow R_1 - 4R_2|\rightarrow\left(\begin{array}{ccc}1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0\end{array}\right),$$

hence $x_{1} + x_{3} = 2, x_{2} = 1$ . Then the solution is $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 \\ 1 \\ 2 - x_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + x_1 \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, x_1 \in \mathbb{R}$ .

It means the system (1) has infinitely many solutions in this case (iii).

Case 2. Let $\alpha = -1$ . Then

$$\left(\begin{array}{ccc}1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & (1 - \alpha) (1 + \alpha)\end{array}\right|_{2(1 - \alpha)}^{6}\rightarrow\left(\begin{array}{ccc}1 & 4 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right)$$

The third row yields $0 = 4$ , which is false. It means the system (1) has no solution in this case (ii).

Case 3. Let $\alpha \neq 1$ and $\alpha \neq -1$. Then

\begin{array}{l} \left( \begin{array}{cccc} 1 & 4 & \alpha & 6 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & (1 - \alpha)(1 + \alpha) & 2(1 - \alpha) \end{array} \right) \to \left| R_3 \leftarrow \frac{R_3}{(1 - \alpha)(1 + \alpha)} \right| \to \left( \begin{array}{ccc} 1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \begin{array}{c} 6 \\ 1 \\ \hline 2(1 - \alpha) \\ \hline ((1 - \alpha)(1 + \alpha)) \end{array} \right) \to \\ \to \left( \begin{array}{ccc} 1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \to \left| R_1 \leftarrow R_1 - 4R_2 \right| \to \\ \to \left( \begin{array}{ccc} 1 & 0 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \to \left| R_1 \leftarrow R_1 - \alpha R_3 \right| \to \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \begin{array}{c} 2 - 2\alpha/(1 + \alpha) \\ 1 \\ 2/(1 + \alpha) \end{array} \right) \to \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \end{array} \end{array} \to \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)

hence $x_1 = \frac{2}{1 + \alpha}$, $x_2 = 1$, $x_3 = \frac{2}{1 + \alpha}$. It means the system (1) has the unique solution in this case (i).

Answer:

(i) $\alpha \neq 1$ and $\alpha \neq -1$;

(ii) $\alpha = -1$;

(iii) $\alpha = 1$.

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