The paragraphs are given below from page 195:
But I do not understand the cases he is describing in 1 through 4, could anyone explain those cases for me by numerical examples, please?
Thanks!
The paragraphs are given below from page 195:
But I do not understand the cases he is describing in 1 through 4, could anyone explain those cases for me by numerical examples, please?
Thanks!
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After we obtain the reduced row echelon form, there are a few possibilities if the corresponding $A'$ and $w'$ are not both zeros, that is if the reduced augmented matrix is not a zero matrix.
We look for the last row of the augemented matrix which is not a zero row then we consider cases by where is the first non-zero position.
Suppose $[A', w']=\left[ \begin{array} {ccc|c} 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]$, this falls into the first case. The last non-zero row of the $A'$ is $[0,0,0|1]$, there is no non-zero entries in the last non-zero row of $A'$. If we consider $A'x = w'$, from the last non-zero row, this means $0=1$, which is a contradiction. We can conclude that there is no solution.
Suppose $[A', w']=\left[ \begin{array} {ccc|c} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{array}\right]$, this falls into the second case. The last non-zero row of the $A'$ is $[0,0,1|1]$. If we consider $A'x = w'$, from the last non-zero row, this means $x_3=1$. Hence, we have solved for $x_3$ and we just have to solve for $x_1$ and $x_2$.
Suppose $[A', w']=\left[ \begin{array} {ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0\end{array}\right]$, this falls into the third case. The last non-zero row of the $A'$ is $[0,1,1|1]$. If we consider $A'x = w'$, from the last non-zero row, this means $x_2+x_3=1$. Hence, we have $x_2=1-x_3$ and we will express $x_2$ in terms of $x_3$. Here $x_3$ is a free parameter. We just have to solve for $x_1$.
If we fall into the first case, we are done.
If we fall into the second or third case, now you can move on to the second last non-zero row and repeat the procedure to solve for the remaining unsolved variables. Possibly in terms of some free variables. This is considered as the $4$-th point.