Answer on Question #60834 – Math – Linear Algebra
Question
a) The Gauss elimination method is used to solve the system of equations
6 × 4 × 1 + 2 + α 3 = 3 × 2 × 1 − 2 + α 3 = 5 × 3 × α × 1 + 2 + 3 = \begin{array}{l}
6 \times 4 \times 1 + 2 + \alpha 3 = \\
3 \times 2 \times 1 - 2 + \alpha 3 = \\
5 \times 3 \times \alpha \times 1 + 2 + 3 = \\
\end{array} 6 × 4 × 1 + 2 + α 3 = 3 × 2 × 1 − 2 + α 3 = 5 × 3 × α × 1 + 2 + 3 =
Find the value of α \alpha α
(i) a unique solution;
(ii) no solution;
(iii) infinitely many solutions.
Solution
We have the system of equation
{ x 1 + 4 x 2 + α x 3 = 6 , 2 x 1 − x 2 + 2 α x 3 = 3 , α x 1 + 3 x 2 + x 3 = 5. \left\{
\begin{array}{l}
x_1 + 4x_2 + \alpha x_3 = 6, \\
2x_1 - x_2 + 2\alpha x_3 = 3, \\
\alpha x_1 + 3x_2 + x_3 = 5.
\end{array}
\right. ⎩ ⎨ ⎧ x 1 + 4 x 2 + α x 3 = 6 , 2 x 1 − x 2 + 2 α x 3 = 3 , α x 1 + 3 x 2 + x 3 = 5.
or in the form A x = b Ax = b A x = b
( 1 4 α 2 − 1 2 α α 3 1 ) ( x 1 x 2 x 3 ) = ( 6 3 5 ) . \left(
\begin{array}{ccc}
1 & 4 & \alpha \\
2 & -1 & 2\alpha \\
\alpha & 3 & 1
\end{array}
\right)
\left(
\begin{array}{c}
x_1 \\
x_2 \\
x_3
\end{array}
\right)
=
\left(
\begin{array}{c}
6 \\
3 \\
5
\end{array}
\right). ⎝ ⎛ 1 2 α 4 − 1 3 α 2 α 1 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 6 3 5 ⎠ ⎞ .
We shall use the Gauss elimination method to solve the system. So put the augmented matrix into the upper triangular form. Firstly, subtract the first row multiplied by 2 from the second row and divide the second row by − 9 -9 − 9
( 1 4 α 2 − 1 2 α α 3 1 ) ⇒ ( 1 4 α 0 − 9 0 α 3 1 ) ⇒ ( 1 4 α 0 1 0 α 3 1 ) ⇒ ( 1 4 α 0 1 0 0 6 α ) → ( 1 4 α 0 1 0 0 4 α − 3 α 2 − 1 ) ⇒ ( 6 6 1 6 α − 5 ) . \left(
\begin{array}{ccc}
1 & 4 & \alpha \\
2 & -1 & 2\alpha \\
\alpha & 3 & 1
\end{array}
\right)
\Rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & -9 & 0 \\
\alpha & 3 & 1
\end{array}
\right)
\Rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
\alpha & 3 & 1
\end{array}
\right)
\Rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
0 & 6\alpha
\end{array}
\right)
\rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
0 & 4\alpha - 3 & \alpha^2 - 1
\end{array}
\right)
\Rightarrow
\left(
\begin{array}{ccc}
6 & 6 \\
1 \\
6\alpha - 5
\end{array}
\right). ⎝ ⎛ 1 2 α 4 − 1 3 α 2 α 1 ⎠ ⎞ ⇒ ⎝ ⎛ 1 0 α 4 − 9 3 α 0 1 ⎠ ⎞ ⇒ ⎝ ⎛ 1 0 α 4 1 3 α 0 1 ⎠ ⎞ ⇒ ⎝ ⎛ 1 0 0 4 1 6 α α 0 ⎠ ⎞ → ⎝ ⎛ 1 0 0 4 1 4 α − 3 α 0 α 2 − 1 ⎠ ⎞ ⇒ ⎝ ⎛ 6 1 6 α − 5 6 ⎠ ⎞ .
Then subtract the first row multiplied by α \alpha α ( − 1 ) (-1) ( − 1 )
( 1 4 α 0 1 0 α 3 1 ) → ( 1 4 α 0 1 0 0 3 − 4 α 1 − α 2 ) → ( 1 4 α 0 1 0 0 4 α − 3 α 2 − 1 ) → ( 1 6 6 1 2 α − 2 ) . \left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
\alpha & 3 & 1
\end{array}
\right)
\rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
0 & 3 - 4\alpha & 1 - \alpha^2
\end{array}
\right)
\rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
0 & 4\alpha - 3 & \alpha^2 - 1
\end{array}
\right)
\rightarrow
\left(
\begin{array}{ccc}
1 & 6 & 6 \\
1 & 2\alpha - 2
\end{array}
\right). ⎝ ⎛ 1 0 α 4 1 3 α 0 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 4 1 3 − 4 α α 0 1 − α 2 ⎠ ⎞ → ⎝ ⎛ 1 0 0 4 1 4 α − 3 α 0 α 2 − 1 ⎠ ⎞ → ( 1 1 6 2 α − 2 6 ) .
In the next step we subtract the second row multiplied by ( 4 α − 3 ) (4\alpha - 3) ( 4 α − 3 )
( 1 4 α 0 1 0 0 4 α − 3 α 2 − 1 ) → ( 1 4 α 0 1 0 0 0 α 2 − 1 ) → ( 6 6 1 1 2 α − 2 ) . \left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
0 & 4\alpha - 3 & \alpha^2 - 1
\end{array}
\right)
\rightarrow
\left(
\begin{array}{ccc}
1 & 4 & \alpha \\
0 & 1 & 0 \\
0 & 0 & \alpha^2 - 1
\end{array}
\right)
\rightarrow
\left(
\begin{array}{ccc}
6 & 6 & 1 \\
1 & 2\alpha - 2
\end{array}
\right). ⎝ ⎛ 1 0 0 4 1 4 α − 3 α 0 α 2 − 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 4 1 0 α 0 α 2 − 1 ⎠ ⎞ → ( 6 1 6 2 α − 2 1 ) .
Then subtract the second row multiplied by 4 from the first row:
\left( \begin{array}{ccc} 1 & 4 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & \alpha^2 - 1 \end{array} \right| \quad \begin{array}{c} 6 \\ 1 \\ 2\alpha - 2 \end{array} \rightarrow \left( \begin{array}{ccc} 1 & 0 & \alpha \\ 0 & 1 & 0 \\ 0 & 0 & \alpha^2 - 1 \end{array} \right| \quad \begin{array}{c} 2 \\ 1 \\ 2\alpha - 2 \end{array} \).
Now we need to find the value of α \alpha α
From the third row it follows that
x 3 = 2 α − 2 α 2 − 1 = 2 α + 1 , α ≠ ± 1. x_3 = \frac{2\alpha - 2}{\alpha^2 - 1} = \frac{2}{\alpha + 1}, \quad \alpha \neq \pm 1. x 3 = α 2 − 1 2 α − 2 = α + 1 2 , α = ± 1.
From the second row it follows that
x 2 = 1. x_2 = 1. x 2 = 1.
From the first row it follows that
x 1 + 2 α α + 1 = 2 ; x_1 + \frac{2\alpha}{\alpha + 1} = 2; x 1 + α + 1 2 α = 2 ; x 1 = 2 − 2 α α + 1 = 2 α + 2 − 2 α α + 1 = 2 α + 1 . x_1 = 2 - \frac{2\alpha}{\alpha + 1} = \frac{2\alpha + 2 - 2\alpha}{\alpha + 1} = \frac{2}{\alpha + 1}. x 1 = 2 − α + 1 2 α = α + 1 2 α + 2 − 2 α = α + 1 2 .
So if α ≠ ± 1 \alpha \neq \pm 1 α = ± 1
(iii) If α = 1 \alpha = 1 α = 1
\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right| \quad \begin{array}{c} 2 \\ 1 \\ 0 \end{array} \).
In other words, we have only two equations for three variables. In this case the system has infinitely many solutions.
(ii) If α = − 1 \alpha = -1 α = − 1 0 = − 4 0 = -4 0 = − 4
Answer: (i) α ∈ R ∖ { − 1 , 1 } \alpha \in \mathbb{R} \setminus \{-1, 1\} α ∈ R ∖ { − 1 , 1 } α = − 1 \alpha = -1 α = − 1 α = 1 \alpha = 1 α = 1
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