MHB Determine the solution set of the system using the echelon form

[mathmari]

Hey!

Let $\displaystyle{a:=\begin{pmatrix}2 & 1 & 0 & 5 \\ 1 & 0 & 1 & 1 \\ 4 & 1 &2 & 7\end{pmatrix}\in \mathbb{R}^{3\times4}}$ and $\displaystyle{b_1:=\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} , \ b_2:=\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix} \in \mathbb{R}^3}$.

I applied the Gauss algorithm to get the echelon form of the matrix $a$ :
\begin{align*}\begin{pmatrix}2 & 1 & 0 & 5 \\ 1 & 0 & 1 & 1 \\ 4 & 1 &2 & 7\end{pmatrix} & \ \overset{R_2:R_2-\frac{1}{2}\cdot R_1}{\longrightarrow} \ \begin{pmatrix}2 & 1 & 0 & 5 \\ 0 & -\frac{1}{2} & 1 & -\frac{3}{2} \\ 4 & 1 &2 & 7\end{pmatrix} \\ &\ \overset{R_3:R_2-2\cdot R_1}{\longrightarrow} \ \begin{pmatrix}2 & 1 & 0 & 5 \\ 0 & -\frac{1}{2} & 1 & -\frac{3}{2} \\ 0 & -1 &2 & -3\end{pmatrix}\\ &\ \overset{R_3:R_2-2\cdot R_2}{\longrightarrow} \ \begin{pmatrix}2 & 1 & 0 & 5 \\ 0 & -\frac{1}{2} & 1 & -\frac{3}{2} \\ 0 & 0 &0 & 0\end{pmatrix}\end{align*}

Then I want to determine the solution set of the system $ax=b_i$ using the echelon form for $i=1$ and $i=2$.

Does this mean that we can use the echelon form of $a$ to calculate the solution or do we use the echelon form of the extended matrix $(a\mid b_i)$ ? (Wondering)

 

[I like Serena]

Then I want to determine the solution set of the system $ax=b_i$ using the echelon form for $i=1$ and $i=2$.

Does this mean that we can use the echelon form of $a$ to calculate the solution or do we use the echelon form of the extended matrix $(a\mid b_i)$ ?

Hey mathmari!

I'm afraid that we'll have to use the echelon form of the extended matrix $(a\mid b_i)$. (Thinking)

Alternatively we could use the echelon form of the extended matrix $(a\mid I_3)$ where $I_3$ is the 3x3 identity matrix. (Thinking)

 

[HallsofIvy]

You can do both problems at once by row reducing
\begin{pmatrix}2 & 1 & 0 & 5 & 1 & -2 \\ 1 & 0 & 1 & 1 & 1 & 1 \\ 4 & 1 & 2 & 7 & 1 & 0\end{pmatrix} where the last two columns are b_1 and b_2.

Klas Van Aarsen's suggestion that you row reduce

\begin{pmatrix}2 & 1 & 0 & 5 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0\\ 4 & 1 & 2 & 7 & 0 & 0 & 1 \end{pmatrix}
uses the fact that once you have row reduced A, the same operations will have converted the identity matrix to A^{-1}.
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