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- Apr 14, 2013

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Hey!!

We have the matrix $A=\begin{pmatrix}1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{pmatrix}$ and the vectors $b_1=\begin{pmatrix}1 \\ 0 \\1\end{pmatrix}$ and $b_2=\begin{pmatrix}-1 \\ 1 \\2\end{pmatrix}$.

Check if the system $Ax=b_i$ for $i\in \{1,2\}$ has a solution.

If the system is impossible find the solution that we get if the vector $b_i$ is projected onto the column space.

If the system has infinitely many solutions find the solution that belongs to the row space.

I have done the following:

We have the matrix $A=\begin{pmatrix}1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{pmatrix}$ and the vectors $b_1=\begin{pmatrix}1 \\ 0 \\1\end{pmatrix}$ and $b_2=\begin{pmatrix}-1 \\ 1 \\2\end{pmatrix}$.

Check if the system $Ax=b_i$ for $i\in \{1,2\}$ has a solution.

If the system is impossible find the solution that we get if the vector $b_i$ is projected onto the column space.

If the system has infinitely many solutions find the solution that belongs to the row space.

I have done the following:

- For $b_1$ the echelon form of the extended matrix is $\begin{pmatrix}\left.\begin{matrix}

\begin{matrix}1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}

\end{matrix}\right|\begin{matrix}1 \\ -1 \\ 2\end{matrix}\end{pmatrix}$.

That means that the system is impossible.

A basis for the column space of $A$ is $\left \{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}, \begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}\right \}$, right?

Let them be $q_1$ and $q_2$ respectively.

The projection of $b$ onto the column space of $A$ is \begin{equation*}b_{p1}=(b_1^Tq_1)q_1+(b_1^Tq_2)q_2=1\cdot q_1+0\cdot g_2=q_1\end{equation*}

Applying again the Gaussian elimination method with the new vector we get the echelon form $\begin{pmatrix}\left.\begin{matrix}

\begin{matrix}1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}

\end{matrix}\right|\begin{matrix}1 \\ 0 \\ 0\end{matrix}\end{pmatrix}$.

So we get the solution \begin{equation*}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}1+x_2 \\ x_2\\ -x_2\end{pmatrix}=\begin{pmatrix}1 \\ 0\\ 0\end{pmatrix}+x_2\begin{pmatrix}1 \\ 1\\ -1\end{pmatrix}\end{equation*}

Is everything correct?

- For $b_2$ the echelon form of the extended matrix is $\begin{pmatrix}\left.\begin{matrix}

\begin{matrix}1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}

\end{matrix}\right|\begin{matrix}-1 \\ 2 \\ 0\end{matrix}\end{pmatrix}$.

That means that the system has infinitely many solutions.

A basis for the row space of $A$ is $\left \{\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}, \begin{pmatrix}0\\ 1\\ 1\end{pmatrix}\right \}$, right?

We get the solution \begin{equation*}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}x_1 \\ x_1+1\\ 1-x_1\end{pmatrix}=\begin{pmatrix}0 \\ 1\\ 1\end{pmatrix}+x_1\begin{pmatrix}1 \\ 1\\ -1\end{pmatrix}\end{equation*}

How do we see that this belongs to the row space? Do we have to check if it is a linear combination of the basis vectors of the row space?

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