This note follows Gilbert Strang's Section 3.3, The Complete Solution to \(Ax=b\), and preserves the questions that helped make the structure intuitive.
1. Why \(Ax=b\) is different from \(Ax=0\)
For
\[ Ax=0, \]
the right side remains zero during elimination.
But for
\[ Ax=b, \]
row operations must also be applied to \(b\). Therefore we work with the augmented matrix
\[ [A\ b]. \]
Elimination gives
\[ [A\ b]\longrightarrow[R\ d], \]
and the systems
\[ Ax=b \]
and
\[ Rx=d \]
have exactly the same solutions.
The first question is therefore not only 'what is x?' but also:
\[ \boxed{\text{Does a solution exist at all?}} \]
2. Solvability and zero rows
Consider
\[ A= \begin{bmatrix} 1&3&0&2\\ 0&0&1&4\\ 1&3&1&6 \end{bmatrix}, \qquad b= \begin{bmatrix} 1\\ 6\\ 7 \end{bmatrix}. \]
Elimination gives
\[ [R\ d] = \left[ \begin{array}{cccc|c} 1&3&0&2&1\\ 0&0&1&4&6\\ 0&0&0&0&0 \end{array} \right]. \]
The last equation is
\[ 0=0. \]
Therefore there is no contradiction and the system is solvable.
Why did this happen?
In the original matrix,
\[ \text{row 3}=\text{row 1}+\text{row 2}. \]
The right side obeys the same relation:
\[ 7=1+6. \]
For a general right side
\[ b= \begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix}, \]
the final equation becomes
\[ 0=b_3-b_1-b_2. \]
Therefore the solvability condition is
\[ \boxed{b_3=b_1+b_2}. \]
The important idea is
\[ \boxed{\text{dependencies among the rows of }A\text{ impose matching conditions on }b.} \]
If elimination instead produced
\[ 0=c,\qquad c\neq0, \]
the system would be impossible.
3. Finding one particular solution
After elimination,
\[ R= \begin{bmatrix} 1&3&0&2\\ 0&0&1&4\\ 0&0&0&0 \end{bmatrix}, \qquad d= \begin{bmatrix} 1\\ 6\\ 0 \end{bmatrix}. \]
The pivot variables are
\[ x_1,\;x_3, \]
while
\[ x_2,\;x_4 \]
are free.
To obtain one convenient particular solution, set the free variables to zero:
\[ x_2=0,\qquad x_4=0. \]
Then
\[ x_1=1,\qquad x_3=6. \]
Therefore
\[ \boxed{ x_p= \begin{bmatrix} 1\\ 0\\ 6\\ 0 \end{bmatrix} }. \]
The free variables do not have to be zero. Zero is chosen only because it gives an easy particular solution.
4. Adding the nullspace gives every solution
Now solve
\[ Rx=0. \]
The two free variables produce the special solutions
\[ s_1= \begin{bmatrix} -3\\ 1\\ 0\\ 0 \end{bmatrix}, \qquad s_2= \begin{bmatrix} -2\\ 0\\ -4\\ 1 \end{bmatrix}. \]
Therefore
\[ x_n=c_1s_1+c_2s_2. \]
The complete solution is
\[ \boxed{x=x_p+x_n}. \]
Explicitly,
\[ \boxed{ x= \begin{bmatrix} 1\\ 0\\ 6\\ 0 \end{bmatrix} +c_1 \begin{bmatrix} -3\\ 1\\ 0\\ 0 \end{bmatrix} +c_2 \begin{bmatrix} -2\\ 0\\ -4\\ 1 \end{bmatrix} }. \]
Why does this work?
\[ A(x_p+x_n)=Ax_p+Ax_n=b+0=b. \]
The particular solution gives one input that produces \(b\). Nullspace vectors are directions invisible to \(A\), so adding them does not change the output.
5. Why a square invertible matrix has \(m=n=r\)
The symbols mean
\[ m=\text{number of rows}, \qquad n=\text{number of columns}, \qquad r=\text{rank}. \]
A square matrix has
\[ m=n. \]
If the square matrix is invertible, elimination reduces it all the way to the identity matrix:
\[ A\longrightarrow I. \]
The identity matrix has a pivot in every row and every column.
Therefore an invertible \(n\times n\) matrix has \(n\) pivots:
\[ r=n. \]
Since it is square,
\[ m=n, \]
so altogether
\[ \boxed{m=n=r}. \]
This also means there are no free variables and
\[ N(A)=\{0\}. \]
Therefore
\[ x=x_p+x_n=A^{-1}b+0=A^{-1}b, \]
so the solution is unique.
6. Why invertibility requires a pivot in every column
Suppose an invertible square matrix were missing a pivot in one column.
That column would be a free column, so the corresponding variable would be free.
Then \(Ax=0\) would have a nonzero solution.
But an invertible matrix must satisfy
\[ Ax=0\Rightarrow x=0. \]
Therefore an invertible square matrix cannot have a free column.
Hence every column has a pivot.
Because the matrix is square, the same number of pivots also fills every row.
So
\[ \boxed{\text{invertible}\Rightarrow\text{pivot in every row and every column}.} \]
7. Full column rank
Full column rank means
\[ \boxed{r=n}. \]
Every column has a pivot.
Therefore there are no free variables and
\[ N(A)=\{0\}. \]
If \(Ax=b\) has two solutions \(x_1,x_2\), then
\[ Ax_1=b, \qquad Ax_2=b. \]
Subtracting gives
\[ A(x_1-x_2)=0. \]
Since the nullspace contains only zero,
\[ x_1-x_2=0, \]
so
\[ x_1=x_2. \]
Therefore full column rank gives
\[ \boxed{\text{at most one solution}.} \]
But a tall matrix may contain zero rows after elimination, so some right sides \(b\) may fail the consistency conditions.
Thus
\[ \boxed{r=n<m}. \]
A tall matrix with full column rank can therefore have either no solution or exactly one solution.
8. Full row rank
Full row rank means
\[ \boxed{r=m}. \]
Every row has a pivot. Therefore elimination produces no zero rows, so every \(b\) is solvable.
If \(n>m\), then
\[ n-r=n-m>0, \]
so free variables exist and the system has infinitely many solutions.
9. Full row rank example
Consider
\[ \begin{aligned} x+y+z&=3,\\ x+2y-z&=4. \end{aligned} \]
Then
\[ A= \begin{bmatrix} 1&1&1\\ 1&2&-1 \end{bmatrix}, \qquad m=2, \quad n=3, \quad r=2. \]
Since
\[ r=m, \]
this matrix has full row rank.
Reduction gives
\[ R= \begin{bmatrix} 1&0&3\\ 0&1&-2 \end{bmatrix}. \]
The free variable is \(x_3\).
Set
\[ x_3=0 \]
to obtain one particular solution:
\[ \boxed{ x_p= \begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix} }. \]
Now solve \(Rx=0\). Set
\[ x_3=1. \]
Then
\[ x_1=-3, \qquad x_2=2, \]
so
\[ \boxed{ s= \begin{bmatrix} -3\\ 2\\ 1 \end{bmatrix} }. \]
The complete solution is
\[ \boxed{ x= \begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix} +t \begin{bmatrix} -3\\ 2\\ 1 \end{bmatrix} }. \]
Geometrically, the two equations describe two planes in \(\mathbb R^3\). Their intersection is a line.
The particular solution is one point on that line.
The nullspace vector gives the direction of the line.
The particular solution is not multiplied by an arbitrary constant. Only the nullspace direction gets the free parameter.
10. The four rank cases
All possibilities can now be classified using only
\[ m,\quad n,\quad r. \]
Case 1: \(r=m=n\)
The matrix is square and invertible.
There are no zero rows and no free variables.
\[ \boxed{Ax=b\text{ has exactly }1\text{ solution}.} \]
Case 2: \(r=m<n\)
The matrix has full row rank and is short and wide.
Every \(b\) is solvable, but free variables exist.
\[ \boxed{Ax=b\text{ has infinitely many solutions}.} \]
Case 3: \(r=n<m\)
The matrix has full column rank and is tall and thin.
There are no free variables, but consistency conditions may fail.
\[ \boxed{Ax=b\text{ has }0\text{ or }1\text{ solution}.} \]
Case 4: \(r<m\) and \(r<n\)
The matrix has neither full row rank nor full column rank.
Zero rows make inconsistency possible, while free variables give infinitely many solutions whenever the system is consistent.
\[ \boxed{Ax=b\text{ has }0\text{ or infinitely many solutions}.} \]
11. The shortest way to remember the four cases
Ask two questions.
Question 1
\[ \boxed{r=m?} \]
If yes, there are no zero rows, so every \(b\) is solvable.
Thus \(r=m\) controls existence for every right side.
Question 2
\[ \boxed{r=n?} \]
If yes, there are no free variables, so there can be at most one solution.
Thus \(r=n\) controls uniqueness.
Those two questions generate all four cases.
Core mental model
The structure can be remembered as
\[ \boxed{\text{zero rows control existence}} \]
because zero rows create consistency conditions on \(b\), while
\[ \boxed{\text{free variables control multiplicity}} \]
because free variables generate nullspace directions.
Therefore:
- no zero rows means every \(b\) is reachable;
- no free variables means at most one \(x\) reaches a given \(b\);
- zero rows plus free variables means either no solution or infinitely many solutions.
This is the structural meaning behind
\[ \boxed{x=x_p+x_n}. \]
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, 5th Edition — Section 3.3: The Complete Solution to Ax = b