Answer: C is $False$
Given:
$Ax=b$ (System of Linear Equations)
Where:
$A = \begin{bmatrix}
a_{11} & ... & a_{1n}\\
\vdots &\vdots &\vdots\\
a_{m1} & ... & a_{mn}
\end{bmatrix}_{m\times n}$
$b=\begin{bmatrix}
b_1\\
\vdots \\
b_m
\end{bmatrix}_{m \times 1}$
$X=\begin{bmatrix}
x_1\\
\vdots \\
x_n
\end{bmatrix}_{n \times 1}$
We will go through each option one-by-one.
A. The system has a solution if and only if, both $A$ and the augmented matrix $[A|b]$ have the same rank:
$=>$ if $Rank(A) = Rank(A|b)$, it means that $b$ can be interpreted as Linear combination of $A$
$=>$ Therefore Solution exists.
$=>$ $True$
B. If $m<n$ and $b$ is the zero vector, then the system has infinitely many solutions.
$=>$ Assume $m=2$ and $n=3$
$=>$ $\begin{bmatrix}
a_{11} & a_{12} & a_{13} & : 0\\
a_{21} & a_{22} & a_{23} & : 0
\end{bmatrix}$
$=>$ max rank possible is always $min(m,n) = min(2,3) = 2$
$=>$ We got one (atleast) free variable, and this free variable can take infinitely many values
$=>$ System has infinitely many solutions
$=>$ $True$
C. If $m=n$ and $b$ is a non-zero vector, then the system has a unique solution.
$=>$ We can’t say really as it might be possible that one column or row is dependent on the other
$=>$ Example, assume matrix of $m=n=3$
$=>$ $\begin{bmatrix}
a_{11} & 2a_{11} & a_{13} & :b_1\\
a_{21} & 2a_{21} & a_{23} & :b_2\\
a_{31} & 2a_{31} & a_{33} & :b_3
\end{bmatrix}$
$=>$ Even if we have $m=n$ here, we can clearly say that $column_2$ is just $2\times column_1$
$=>$ No unique solution exists
$=>$ $False$
D. The system will have only a trivial solution when $m=n$, $b$ is the zero vector and $rank(A)=n$.
$=>$ Since, $rank(A)=n$, all the columns are having pivot
$=>$ All columns are having pivot means all columns are Linearly Independent
$=>$ Thus only possible solution to get a zero vector $b$ would be to multiply each column with $0$ in $A$, no other possibility
$=>$ When we multiply every column with $zero$, it becomes $Trivial$ $Solution$
$=>$ $True$