MHB La1.1.22 3 aug matrices for linear sys solution set is x_1=3,x_2=-2,x_3=-1

karush
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$$\tiny{la1.1.22}$$

$\textsf{Construct 3 different augmented matrices for linear systems whose solution set is}$
$$x_1=3,\quad x_2=-2,\quad x_3=-1$$Well we could start just by$$3x-2y+y=-1$$but then we need a $3\times4$ matrix
 
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karush said:
$$\tiny{la1.1.22}$$

$\textsf{Construct 3 different augmented matrices for linear systems whose solution set is}$
$$x_1=3,\quad x_2=-2,\quad x_3=-1$$

Well we could start just by
$$3x-2y+y=-1$$

When we substitute $x=3$ and $y=-2$, we get $3\cdot 3 - 2\cdot (-2) + (-2)=11 \ne -1$.
That's not going to work is it?

karush said:
but then we need a $3\times4$ matrix

We need indeed a 3x4 matrix since we have 3 unknowns, meaning we should have 3 equations for them.
The simplest augmented matrix would follow from:
$$\begin{cases}x_1=3 \\ x_2=-2 \\ x_3=-1\end{cases}$$
What is the corresponding augmented matrix?
 
I like Serena said:
When we substitute $x=3$ and $y=-2$, we get $3\cdot 3 - 2\cdot (-2) + (-2)=11 \ne -1$.
That's not going to work is it?

well actually its $x+y+z=3-2-1=0$
I like Serena said:
We need indeed a 3x4 matrix since we have 3 unknowns, meaning we should have 3 equations for them.
The simplest augmented matrix would follow from:
$$\begin{cases}x_1=3 \\ x_2=-2 \\ x_3=-1\end{cases}$$
What is the corresponding augmented matrix?

ok presumably this would be one case

$$\left[\begin{array}{rrrrr}
%x_1& x_2& x_3& 0\\
x_1& 0& 0& 3\\
0& x_2& 0& -2\\
0& 0& x_3& -1\\
\end{array}\right]$$
 
Last edited:
karush said:
ok presumably this would be one case

$$\left[\begin{array}{rrrrr}
%x_1& x_2& x_3& 0\\
x_1& 0& 0& 3\\
0& x_2& 0& -2\\
0& 0& x_3& -1\\
\end{array}\right]$$

... except that an augmented matrix is without the variables.
So it should be:
$$\left[\begin{array}{rrrrr}
%x_1& x_2& x_3& 0\\
1& 0& 0& 3\\
0& 1& 0& -2\\
0& 0& 1& -1\\
\end{array}\right]$$

And if we add both the second row and the third row to the first row, the first row becomes the equation that you've suggested... and we have a new augmented matrix.
 
I like Serena said:
... except that an augmented matrix is without the variables.
So it should be:
$$\left[\begin{array}{rrrrr}
%x_1& x_2& x_3& 0\\
1& 0& 0& 3\\
0& 1& 0& -2\\
0& 0& 1& -1\\
\end{array}\right]$$

And if we add both the second row and the third row to the first row, the first row becomes the equation that you've suggested... and we have a new augmented matrix.

the only other augmented matrix I see is just flipping this one to$$\left[\begin{array}{rrrrr}
0& 0& 1& -1\\
0& 1& 0& -2\\
1& 0& 0& 3\\
\end{array}\right]$$so the the third ?
 
karush said:
the only other augmented matrix I see is just flipping this one to$$\left[\begin{array}{rrrrr}
0& 0& 1& -1\\
0& 1& 0& -2\\
1& 0& 0& 3\\
\end{array}\right]$$so the the third ?

Do the reverse of a row reduction?
That is, add/substract rows such that instead of elements becoming zero, they become non-zero?
 
I like Serena said:
Do the reverse of a row reduction?
That is, add/substract rows such that instead of elements becoming zero, they become non-zero?
i thot augmented meant you had to have a triangle of zeros
 
karush said:
i thot augmented meant you had to have a triangle of zeros

Neh, an augmented matrix is just an encoding of a set of equations into a single matrix.
When we apply row reduction to an augmented matrix, we get a new augmented matrix in row reduced echelon form, which has indeed a triangle of zeroes.
 
I like Serena said:
Neh, an augmented matrix is just an encoding of a set of equations into a single matrix.
When we apply row reduction to an augmented matrix, we get a new augmented matrix in row reduced echelon form, which has indeed a triangle of zeroes.

ok here goes...

$$A=\left[\begin{array}{rrrrr}
0& 0& 1& -1\\ 0& 1& 0& -2\\ 1& 0& 0& 3\\
\end{array}\right]$$
$\textsf{so then if from $A$ we multiply $R_1$ by $2$ and add to $R_3$ to become $A_2$}$
$$A_2=\left[\begin{array}{rrrrr}
0& 0& 2& -2\\ 0& 1& 0& -2\\ 1& 0& 2& 1\\
\end{array}\right]$$
$\textsf{so then if from $A$ we multiply $R_1$ by $2$ and add to $R_2$ to become $A_3$}$
$$A_3=\left[\begin{array}{rrrrr}
0& 0& 2& -2\\ 0& 1& 2& 0\\ 1& 0& 0& 3\\
\end{array}\right]$$so then thusly
kinda maybe hopefully?
 
  • #10
karush said:
ok here goes...

$$A=\left[\begin{array}{rrrrr}
0& 0& 1& -1\\ 0& 1& 0& -2\\ 1& 0& 0& 3\\
\end{array}\right]$$
$\textsf{so then if from $A$ we multiply $R_1$ by $2$ and add to $R_3$ to become $A_2$}$
$$A_2=\left[\begin{array}{rrrrr}
0& 0& 2& -2\\ 0& 1& 0& -2\\ 1& 0& 2& 1\\
\end{array}\right]$$
$\textsf{so then if from $A$ we multiply $R_1$ by $2$ and add to $R_2$ to become $A_3$}$
$$A_3=\left[\begin{array}{rrrrr}
0& 0& 2& -2\\ 0& 1& 2& 0\\ 1& 0& 0& 3\\
\end{array}\right]$$so then thusly
kinda maybe hopefully?

All good except for $A_3$ where there is a mistake with a sign.
 

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