Option A:
We know that
- row rank (A) = column rank (A) = rank(A)
- row rank (A) <= m
- column rank (A) <= n
Given $n < m$ . So, $rank (A) <= n$.
Note that Augmented matrix (A | b) has m rows and n+1 column.
So, $rank(A|b) <= n+1$. We already know that $n<m \equiv n+1 <= m$, hence we can say $rank(A|b) <= n+1 <=m$.
Therefore, option A is True.
Option B:
(Copied Suraj’s comment)
Note: For a matrix $A_{m \times n}$, when the rank equals the smallest dimension (i.e., min(m, n)), it is said to have full rank.
Consider $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} x_1 \end{bmatrix} $ = $\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}$.
This matrix has full rank but does not have a solution.
Therefore, option B is False.
Option C:
Consider $A = \begin{bmatrix} 3 & 6 \\ 2 & 4 \\ 1 & 2 \end{bmatrix}, X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $ and b = $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$.
We see that $AX=b$ has a solution ($x_1=1, x_2=0$) but $AX=0$ has infinitely many solutions.
Therefore, option C is False.
Option D:
(Copied Suraj’s comment)
Consider $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} x_1 \end{bmatrix} $ = $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$.
It has unique solution which is $x_1 =1$ but is not invertible.
Therefore, option D is False.