- #1

mathmari

Gold Member

MHB

- 5,049

- 7

Hey!

We have the vectors $\overrightarrow{a_1}=\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}, \overrightarrow{a_2}=\begin{pmatrix}-1 \\0 \\ 2\end{pmatrix}, \overrightarrow{a_3}=\begin{pmatrix}7 \\ 8 \\ 6\end{pmatrix}$.

I have shown that these vectors are linearly dependent:

$\begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

2 & 0 & 8\\

3 & 2 & 6

\end{matrix}\left|\begin{matrix}

0\\

0\\

0

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

2.\text{row}-2\cdot 1.\text{row}\\

3.\text{row}-3\cdot 1.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 5 & -15

\end{matrix}\left|\begin{matrix}

0\\

0\\

0

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

\\

3.\text{row}-\frac{5}{2}\cdot 2.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 0 & 0

\end{matrix}\left|\begin{matrix}

0\\

0\\

0

\end{matrix}\right.\end{bmatrix}$

right? (Wondering)

Can we write the vectors $\overrightarrow{b}=\begin{pmatrix}-6 \\ -4 \\ 2\end{pmatrix}$ and $\overrightarrow{c}=\begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}$ as linear combinations of the vectors $\overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3}$ ? (Wondering)

To check this I applied the Gauss elimination algorithm:

$\begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

2 & 0 & 8\\

3 & 2 & 6

\end{matrix}\left|\begin{matrix}

-6\\

-4\\

2

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

2.\text{row}-2\cdot 1.\text{row}\\

3.\text{row}-3\cdot 1.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 5 & -15

\end{matrix}\left|\begin{matrix}

-6\\

8 \\

20

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

\\

3.\text{row}-\frac{5}{2}\cdot 2.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 0 & 0

\end{matrix}\left|\begin{matrix}

-6\\

8\\

0

\end{matrix}\right.\end{bmatrix}$ $\begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

2 & 0 & 8\\

3 & 2 & 6

\end{matrix}\left|\begin{matrix}

0\\

2\\

1

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

2.\text{row}-2\cdot 1.\text{row}\\

3.\text{row}-3\cdot 1.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 5 & -15

\end{matrix}\left|\begin{matrix}

0\\

2 \\

3

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

\\

3.\text{row}-\frac{5}{2}\cdot 2.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 0 & 0

\end{matrix}\left|\begin{matrix}

0\\

2\\

-2

\end{matrix}\right.\end{bmatrix}$

So, in both cases the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ cannot be written as linear combinations of the vectors $\overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3}$.

Is everything correct? Could I improve something? (Wondering)

I have to give all the possible solutions of the linear equations system.

For the vector $\overrightarrow{b}$:

we get $2\lambda_2-6\lambda_3=8$ and $\lambda_1-\lambda_2+7\lambda_3=-6$.

Solving at the first equation for $\lambda_2$ we get $\lambda_2=4+3\lambda_3$.

From the other equation we get $\lambda_1=\lambda_2-7\lambda_3-6 =4+3\lambda_3-7\lambda_3-6=-2-4\lambda_3$.

So, are all the possible solutions of the linear equations system: $(\lambda_1, \lambda_2, \lambda_3)=(-2-4\lambda_3, 4+3\lambda_3, \lambda_3)=\lambda_3(-4, 3, 1)+(-2, 4, 0)$ ? (Wondering)

We have the vectors $\overrightarrow{a_1}=\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}, \overrightarrow{a_2}=\begin{pmatrix}-1 \\0 \\ 2\end{pmatrix}, \overrightarrow{a_3}=\begin{pmatrix}7 \\ 8 \\ 6\end{pmatrix}$.

I have shown that these vectors are linearly dependent:

$\begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

2 & 0 & 8\\

3 & 2 & 6

\end{matrix}\left|\begin{matrix}

0\\

0\\

0

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

2.\text{row}-2\cdot 1.\text{row}\\

3.\text{row}-3\cdot 1.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 5 & -15

\end{matrix}\left|\begin{matrix}

0\\

0\\

0

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

\\

3.\text{row}-\frac{5}{2}\cdot 2.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 0 & 0

\end{matrix}\left|\begin{matrix}

0\\

0\\

0

\end{matrix}\right.\end{bmatrix}$

right? (Wondering)

Can we write the vectors $\overrightarrow{b}=\begin{pmatrix}-6 \\ -4 \\ 2\end{pmatrix}$ and $\overrightarrow{c}=\begin{pmatrix}0 \\ 2 \\ 1\end{pmatrix}$ as linear combinations of the vectors $\overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3}$ ? (Wondering)

To check this I applied the Gauss elimination algorithm:

$\begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

2 & 0 & 8\\

3 & 2 & 6

\end{matrix}\left|\begin{matrix}

-6\\

-4\\

2

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

2.\text{row}-2\cdot 1.\text{row}\\

3.\text{row}-3\cdot 1.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 5 & -15

\end{matrix}\left|\begin{matrix}

-6\\

8 \\

20

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

\\

3.\text{row}-\frac{5}{2}\cdot 2.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 0 & 0

\end{matrix}\left|\begin{matrix}

-6\\

8\\

0

\end{matrix}\right.\end{bmatrix}$ $\begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

2 & 0 & 8\\

3 & 2 & 6

\end{matrix}\left|\begin{matrix}

0\\

2\\

1

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

2.\text{row}-2\cdot 1.\text{row}\\

3.\text{row}-3\cdot 1.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 5 & -15

\end{matrix}\left|\begin{matrix}

0\\

2 \\

3

\end{matrix}\right.\end{bmatrix}\begin{matrix}

\\

\\

3.\text{row}-\frac{5}{2}\cdot 2.\text{row}

\end{matrix} \longrightarrow \begin{bmatrix}

\begin{matrix}

1 & -1 & 7\\

0 & 2 & -6\\

0 & 0 & 0

\end{matrix}\left|\begin{matrix}

0\\

2\\

-2

\end{matrix}\right.\end{bmatrix}$

So, in both cases the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ cannot be written as linear combinations of the vectors $\overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3}$.

Is everything correct? Could I improve something? (Wondering)

I have to give all the possible solutions of the linear equations system.

For the vector $\overrightarrow{b}$:

we get $2\lambda_2-6\lambda_3=8$ and $\lambda_1-\lambda_2+7\lambda_3=-6$.

Solving at the first equation for $\lambda_2$ we get $\lambda_2=4+3\lambda_3$.

From the other equation we get $\lambda_1=\lambda_2-7\lambda_3-6 =4+3\lambda_3-7\lambda_3-6=-2-4\lambda_3$.

So, are all the possible solutions of the linear equations system: $(\lambda_1, \lambda_2, \lambda_3)=(-2-4\lambda_3, 4+3\lambda_3, \lambda_3)=\lambda_3(-4, 3, 1)+(-2, 4, 0)$ ? (Wondering)

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