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
$$3x-2y+y=-1$$
karush said:but then we need a $3\times4$ 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?
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?
karush said:ok presumably this would be one case
$$\left[\begin{array}{rrrrr}
%x_1& x_2& x_3& 0\\
x_1& 0& 0& 3\\
0& x_2& 0& -2\\
0& 0& x_3& -1\\
\end{array}\right]$$
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.
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 ?
i thot augmented meant you had to have a triangle of zerosI 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?
karush said:i thot augmented meant you had to have a triangle of zeros
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.
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?
The notation "La1.1.22 3 aug matrices for linear sys solution set is x_1=3,x_2=-2,x_3=-1" refers to a linear system of equations that can be represented using augmented matrices. The solution set for this system is given by x_1=3, x_2=-2, and x_3=-1.
An augmented matrix is a representation of a linear system of equations in matrix form, where the coefficients of the variables are placed in a rectangular matrix and the constants are placed in a column vector on the right side of the matrix.
The solution set for a linear system of equations can be found by performing elementary row operations on the augmented matrix until it is in reduced row echelon form. The resulting matrix will have the solutions to the system of equations on the right side.
A matrix is in reduced row echelon form when it has the following properties: 1) all rows of zeros are at the bottom, 2) the first non-zero entry in each row (called a leading entry) is a 1, 3) each leading entry is the only non-zero entry in its column, and 4) the leading entry of each row is to the right of the leading entry in the row above it.
The solutions in an augmented matrix correspond to the solutions to the linear system of equations by using the values in the rightmost column as the values for the variables in the system. In the example provided, x_1=3, x_2=-2, and x_3=-1 would be the solutions to the system of equations.