IF |A| $\neq$ 0 then the equation has only trivial solution i.e all the variables are 0. hence it is the “only” and unique solution. Hence true.

For the 2^{nd} statement it should have one solution and not infinitely many solution. Hence false

For the last statement let’s take an example: x+y=1. Here we have 2 variables and 1 eq^n, unless and until we don't have one more equation we cannot determine the values of x and y uniquely. Hence true.

Q1 IF B=0 AND DETERMINENT OF A i.e |A| IS NOT EQUAL TO ZERO THEN IT MEANS UNIQUE SOLUTION??

If B=0, it means it is a homogeneous system of equations. And all have the trivial solution i.e. origin. Now if |A| $\neq$ 0, it means that all the vectors are independent and there exists just one solution i.e. the trivial solution. Hence, TRUE.

Q2 IF B NOT EQUAL TO 0 AND DETERMINENT OF A i.e |A| IS NOT EQUAL TO ZERO THEN IT MEANS INFINITE MANY SOLUTION SOLUTION?

If B $\neq$0, it means a non-homogeneous system of equations. If |A| $\neq$ 0, then it means that it has a unique solution as all the vectors are independent and they will intersect at just one unique point. Hence, FALSE.

Q3 IF B IS NOT EQUAL TO 0 AND M<N THEN IT MEANS NO UNIQUE SOLUTION??

Again a non-homogeneous system of equations. If M<n, then definitely Rank(A|b) < number of variables. Now either infinite solutions exists (if Rank(A)==Rank(A|b)) or no solution exists ((if Rank(A)<Rank(A|b)).

In simple terms, if for 3 unknowns i give you 2x+3y+4z=5 and 3x+2y+z=9, then by putting different values of one variable you will get infinite solutions. But if give x+y+z = 1 and x+y+z=2, then obviously this is not possible and hence no solution exists. Hence, TRUE.