Row echelon form : Can the first column contain only zeros?

In summary, the row echelon form of matrix A is: \begin{pmatrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}For question (b), the solutions $x_i$ for $1\leq i\leq 3$ are: $x_1 = \begin{
  • #1
mathmari
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Hey! :giggle:

Let \begin{equation*}A=\begin{pmatrix}0 & -2 & 2 & 0 & 0 & -6 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & 0 & 2 & 7\\ 0 & 3 & -3 & 1 & 2 & 14\end{pmatrix}, \ b_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix} , \ b_2=\begin{pmatrix}-2 \\ 1 \\ -1 \\ 3 \\ 5\end{pmatrix}, \ b_3=\begin{pmatrix}-2 \\ 2 \\ 0 \\ 4 \\ 7\end{pmatrix}\end{equation*}

(a) Determine the row echelon form of $A$.

(b) Calculate for all $1\leq i\leq 3$ with $L(A,b_i)\neq \emptyset$ a solution $x_i$.

(c) Give a basis of $L(A, 0_{\mathbb{R}^5})$.

(d) Give $L(A,b_i)$ for all $1\leq i\leq 3$ using the basis of (c).
For question (a) : At the matrix $A$ the first column contains only zeros. Can the echelon form contain only zeros at the first column? Or do we have to exchange the first column with an other one to get a non-zero element at the upper left position? :unsure:
 
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  • #2
Hey mathmari!

It is fine if the first column - and any other columns as well - will contain only zeroes in the row echelon form. 🤔
 
  • #3
Klaas van Aarsen said:
It is fine if the first column - and any other columns as well - will contain only zeroes in the row echelon form. 🤔

So we have the following?
\begin{align*}\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & -2 & 2 & 0 & 0 & -6 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & 0 & 2 & 7\\ 0 & 3 & -3 & 1 & 2 & 14\end{matrix}
\end{matrix}\right|\begin{matrix}0 & -2 & -2 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 3 & 4\\ 0 & 5 & 7\end{matrix}\end{pmatrix} &\ \overset{R_1:\frac{1}{-2}\cdot R_1}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & 0 & 2 & 7\\ 0 & 3 & -3 & 1 & 2 & 14\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 3 & 4\\ 0 & 5 & 7\end{matrix}\end{pmatrix} \\ & \ \overset{R_4:R_4-R_1}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 2 & 4\\ 0 & 3 & -3 & 1 & 2 & 14\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 2 & 3\\ 0 & 5 & 7\end{matrix}\end{pmatrix} \\ & \ \overset{R_5:R_5-3\cdot R_1}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 2 & 5\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -1 & 0 \\ 0 & 2 & 3\\ 0 & 2 & 4\end{matrix}\end{pmatrix} \\ & \ \overset{R_3:R_3-R_2}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 2 & 4\\ 0 & 0 & 0 & 1 & 2 & 5\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 2 & 3\\ 0 & 2 & 4\end{matrix}\end{pmatrix} \\ & \ \overset{R_4:R_4+R_3}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 2 & 5\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 0 & 1\\ 0 & 2 & 4\end{matrix}\end{pmatrix} \\ & \ \overset{R_5:R_5-R_2}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 2\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 0 & 1\\ 0 & 1 & 2\end{matrix}\end{pmatrix} \\ & \ \overset{R_5:R_5+\frac{1}{2}\cdot R_3}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & -2 & -2 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{matrix}\end{pmatrix} \\ & \ \overset{R_3:\frac{1}{-2}\cdot R_3}{\rightarrow } \
\begin{pmatrix} \left.\begin{matrix}
\begin{matrix}0 & 1 & -1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 1 & 3\\ 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\end{matrix}
\end{matrix}\right|\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 2\\ 0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{matrix}\end{pmatrix}\end{align*}
That means that $L(A, b_3)=\emptyset$ and $L(A, b_1)\neq\emptyset$ and $L(A, b_1)\neq\emptyset$.

Right? :unsure:
 
  • #4
It looks correct to me. (Nod)
 
  • #5
Klaas van Aarsen said:
It looks correct to me. (Nod)

So for question (b) we have the following :

$x_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\in L(A,b_1)$.

Now $x_2$ is not unique, but since we are asked to calculate a solutoion, I suppose we give one of the solutions, right?
Considering $b_2$ we get the following equations : \begin{align*}b-c+3f=&1 \\ d+e+3f=&1 \\ e+2f=&1\end{align*}
So to get one solution we have to take arbirtray values for the free parameters, right?
From last equation we have $e=1-2f$, for $f=0$ we get $e=1$.
From the second equation we getthen $d=0$.
From the first equation we get $b=1+c$ and for $c=0$ we get $b=1$.

So $x_2=\begin{pmatrix}0\\ 1 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}\in L(A,b_2)$.

Is that correct so far?

:unsure: For question (c) we have that $L(A,0_{\mathbb{R}^5})=L(A,b_1)=\left \{\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\right \}$. Is the zero vector the basis? :unsure: For question (d) why do we need (c) can we not just use (b) ? :unsure:
 
  • #6
mathmari said:
So for question (b) we have the following :

$x_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\in L(A,b_1)$.

Now $x_2$ is not unique, but since we are asked to calculate a solution, I suppose we give one of the solutions, right?

That should indeed be fine. It will come back in (c) though. (Smirk)

mathmari said:
So $x_2=\begin{pmatrix}0\\ 1 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}\in L(A,b_2)$.

Is that correct so far?

If I fill it in, I get indeed $b_2$, so it's correct. (Nod)

mathmari said:
For question (c) we have that $L(A,0_{\mathbb{R}^5})=L(A,b_1)=\left \{\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}\right \}$. Is the zero vector the basis?

A basis must be a set of independent vectors that span the desired space.
The zero vector cannot be an independent vector, so it cannot be an element of a basis either. (Shake)

mathmari said:
For question (d) why do we need (c) can we not just use (b) ? :unsure:

We've found the null space in (c).
If we add one of those vectors to a solution, it will again be a solution.
Either way, in (b) we found a single solution. Now we have to find a basis for all solutions.
The question asks to include the basis for the null space in the basis for all solution, instead of coming up with an unrelated basis. 🤔
 
  • #7
Klaas van Aarsen said:
A basis must be a set of independent vectors that span the desired space.
The zero vector cannot be an independent vector, so it cannot be an element of a basis either. (Shake)

So no matter which basis we take we getthe zero vector, but in this case the only solution is the zero vector, or not?
So what basis do we take?

:unsure:
 
  • #8
mathmari said:
So no matter which basis we take we get the zero vector, but in this case the only solution is the zero vector, or not?
So what basis do we take?
We didn't actually solve $Ax_1=b_1=0$ yet, did we? :unsure:
 
  • #9
Klaas van Aarsen said:
We didn't actually solve $Ax_1=b_1=0$ yet, did we? :unsure:

Ah yes!

As previously we get the equations\begin{align*}b-c+3f=&0 \\ d+e+3f=&0 \\ e+2f=&0\end{align*}
From the last equation we get $e=-2f$.
From the second equation we get $d-2f+3f=0 \Rightarrow d+f=0 \Rightarrow d=-f$.
From the first equation we get $b-c+3f=0 \Rightarrow b=c-3f$.

So the solution vector is \begin{equation*}\begin{pmatrix}a\\ b\\ c\\ d\\ e\\ f\end{pmatrix}=\begin{pmatrix}a\\ c-3f\\ c\\ -f\\ -2f\\ f\end{pmatrix}=\begin{pmatrix}a\\ 0\\ 0\\ 0\\ 0\\ 0\end{pmatrix}+\begin{pmatrix}0\\ c\\ c\\ 0\\ 0\\ 0\end{pmatrix}+\begin{pmatrix}0\\ -3f\\ 0\\ -f\\ -2f\\ f\end{pmatrix}=a\begin{pmatrix}1\\ 0\\ 0\\ 0\\ 0\\ 0\end{pmatrix}+c\begin{pmatrix}0\\ 1\\ 1\\ 0\\ 0\\ 0\end{pmatrix}+f\begin{pmatrix}0\\ -3\\ 0\\ -1\\ -2\\ 1\end{pmatrix}\end{equation*}
So is the basis teh set of these three vectors?

:unsure:
 
  • #10
mathmari said:
So is the basis the set of these three vectors?
Yep. (Nod)
 
  • #11
Klaas van Aarsen said:
Yep. (Nod)

So we get $L(a,b_i)=x_i+B$ with $B$ the basis above, right? :unsure:
 
  • #12
mathmari said:
So we get $L(a,b_i)=x_i+B$ with $B$ the basis above, right?
Aren't there more solutions than just the arbitrary $x_i$ that we picked before? :unsure:
 
  • #13
Klaas van Aarsen said:
Aren't there more solutions than just the arbitrary $x_i$ that we picked before? :unsure:

We have that $L(a,b_1)=B$.

As for $b_2$ :
\begin{align*}b-c+3f=&1 \\ d+e+3f=&1 \\ e+2f=&1\end{align*}
We get $e=1-2f$, $d=-f$, $b=1+c-3f$.
But how do we need here $L(a,0)$ ? :unsure:

We have that $L(a, b_3)=\emptyset$.
 
  • #14
mathmari said:
So we get $L(a,b_i)=x_i+B$ with $B$ the basis above, right?
mathmari said:
As for $b_2$ :
\begin{align*}b-c+3f=&1 \\ d+e+3f=&1 \\ e+2f=&1\end{align*}
We get $e=1-2f$, $d=-f$, $b=1+c-3f$.
But how do we need here $L(a,0)$ ?
You were right after all.
The $x_1$ you is a solution and we can add any vector that belongs to the null space.
So the solution is indeed $x_1+B$. 🤔
 
  • #15
Klaas van Aarsen said:
You were right after all.
The $x_1$ you is a solution and we can add any vector that belongs to the null space.
So the solution is indeed $x_1+B$. 🤔

Great! Thanks a lot! (Handshake)
 

Related to Row echelon form : Can the first column contain only zeros?

What is the row echelon form?

The row echelon form is a way of representing a matrix in a simplified form, where each row has more leading zeros than the row above it.

What is the significance of the first column in the row echelon form?

The first column in the row echelon form is important because it represents the pivot column, which is used to perform row operations and simplify the matrix.

Can the first column in the row echelon form contain only zeros?

Yes, the first column in the row echelon form can contain only zeros. This means that the matrix has no pivot column and is already in its simplest form.

What does it mean if the first column in the row echelon form contains only zeros?

If the first column in the row echelon form contains only zeros, it means that the matrix has no pivot column and is already in its simplest form. This also means that the rows are linearly dependent and the matrix is not invertible.

How do I know if a matrix is in row echelon form?

To determine if a matrix is in row echelon form, you can check if each row has more leading zeros than the row above it, and if the pivot columns are to the right of the pivot columns in the row above it.

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