How to deduce the cases of unique, no, and infinitely many solution from row echelon form?

Question

I got a SLE represented by this augmented matrix $$ \begin{bmatrix} 1 & 1 & -k & h \\ 1 & 2 & 1 & 2 \\ -2 & 1 & 1 & 2 \\ \end{bmatrix} $$

I reduced it to this by applying row operation: $$ \begin{bmatrix} 1 & 0 & -k-3/5 & -1+h \\ 0 & 1 & k+1 & -h+2 \\ 0 & 0 & 1 & (5h-25/3)/(-5k-4) \\ \end{bmatrix} $$

I now have 3 "leading ones", how to conclude which values of k and h yield infinite solutions, a unique solution, and no solution?

Note: condition for this to apply is $k≠4/5$

Answer 1

For problems like this, I like to push the unknown coefficients as far to the bottom of the matrix as I can to keep the intermediate computations simpler and avoid having to deal with potential zero divisions along the way. Rearranging rows in this way corresponds to reordering the equations, which obviously doesn’t affect the system’s solution.

Here, we’re lucky and the two unknowns are in the same row, so start the row-reduction from $$\begin{bmatrix}1&2&1&2\\-2&1&1&2\\1&1&-k&h\end{bmatrix}$$ instead. Eventually, you’ll reach $$\begin{bmatrix}1&2&1&2\\0&5&3&6\\0&0&-\left(k+\frac25\right)&h-\frac45\end{bmatrix}.$$ From that we can readily see that there is an infinite number of solutions when $k=-\frac25$ and $h=\frac45$, no solution when $k=-\frac25$ but $h\ne\frac45$, and a unique solution otherwise.

Answer 2

As a first step, you should subtract the first row from the second one and to add to the third rwo twice the first one; you will get$$\begin{bmatrix}1 & 1 & -k & h \\ 0 & 1 & k+1 & 2-h \\ 0 & 3 & 1-2 k & 2 h+2\end{bmatrix}.$$Now, you should subtract the second row from the first one and add to the third row the second one times $-3$, getting$$\begin{bmatrix}1 & 0 & -2 k-1 & 2 (h-1) \\ 0 & 1 & k+1 & 2-h \\ 0 & 0 & -5 k-2 & 5 h-4\end{bmatrix}.\tag1$$And now there are two cases to be considered. The first case is the case in which $k=-\frac25$; then $(1)$ is equal to$$\begin{bmatrix}1 & 0 & -\frac{1}{5} & 2 (h-1) \\ 0 & 1 & \frac{3}{5} & 2-h \\ 0 & 0 & 0 & 5 h-4\end{bmatrix}.$$Otherwise, divide the third row of $(1)$ by $-5k-2$; you will get$$\begin{bmatrix}1 & 0 & -2 k-1 & 2 (h-1) \\ 0 & 1 & k+1 & 2-h \\ 0 & 0 & 1 & -\frac{5h-4}{5k+2}&\end{bmatrix}.$$Can you take it from here?

Answer 3

The rank of a matrix can be defined in various equivalent ways-in this context the most convenient way to define the rank of a matrix is as the number of leading 1's in its row-reduced echelon form. When the augmented matrix has been brought to r.r.e.form, the part without the last column is the r.r.e. form for the coefficient matrix. The rank, i.e. the number of leading 1's s of the augmented matrix, is the same as or one more than the rank of the coefficient matrix. It is one more iff the r.r.e. form of the augmented matrix has a leading 1 in the last column iff the system is inconsistent. (Not a surprising result-this means that if we have $n$ unknowns, the last equation is $$0x_1+...+0x_n=1$$ which is impossible.) The case that the coefficient and augmented matrix have the same rank, i.e. that the last column does not have a leading 1, splits into two sub-cases (i) the number of leading 1's equals the number of unknowns-then the solution is unique and can be read off directly from the r.r.e. form (ii) the number of leading 1's is less than the number of unknowns. Then the variables corresponding to columns in the r.r.e. form of the coefficient matrix not containing a leding 1 can be given arbitrary values and the system can be solved for the remining variables, i.e. the variables corresponding to the columns in the r.r.e. form of the coefficient matrix containing a leading 1. In this situation there are multiple solutions-indeed, there are infinitely many solutions.