MHB Can the vectors be written as a linear combination?

[mathmari]

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)

 

[HallsofIvy]

Yes, these vectors are dependent. In showing whether or not vector \vec{b} can be written as a linear combination, you arrived at a reduced matrix in which all 4 entries in the last row equal to 0. since 0x+ 0y+ 0z= 0 for any x, y, z, that shows that \vec{b} can be written as a linear combination, in infinitely many ways.

For \vec{c} you arrived at a reduce matrix with last row "0 0 0 | -2". Since 0x+ 0y+ 0z is not equal to -2 or any x, y, z, yes, \vec{c} cannot be written as a linear combination.

 

[mathmari]

Yes, these vectors are dependent. In showing whether or not vector \vec{b} can be written as a linear combination, you arrived at a reduced matrix in which all 4 entries in the last row equal to 0. since 0x+ 0y+ 0z= 0 for any x, y, z, that shows that \vec{b} can be written as a linear combination, in infinitely many ways.

For \vec{c} you arrived at a reduce matrix with last row "0 0 0 | -2". Since 0x+ 0y+ 0z is not equal to -2 or any x, y, z, yes, \vec{c} cannot be written as a linear combination.

I understand! (Smile)

And are the infinitely many ways for $\vec{b}$ described as follows?

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)

 

[HallsofIvy]

Yes. Since your matrix can be row reduced a matrix with one row all 0s, this is equivalent two equations in three variables, a, b, and c. You can solve for two of the variables in terms of the third, or solve for all three in terms of a new parameter. Here you have the single parameter \lambda_3.

 

[I like Serena]

Could I improve something? (Wondering)

Hey mathmari! (Smile)

I would recommend using column elimination instead of row elimination.
As a result we'll find the minimum set of independent vectors that also makes it easier to find the other dependencies. (Nerd)

 

[mathmari]

Ah ok...

Thank you very much! (Happy)