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consider a system of linear equation where  AMxN  XNx1 =BMx1

TRUE OR FALSE

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

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??

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

Q1 ) false

Q2) false

Q3) false
any reason becz answer for 1 nd 3 is given true
edited by

Here’s the thing:

1. 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.
2. For the 2nd statement it should have one solution and not infinitely many solution. Hence false
3. 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.

isn’t 2nd statement for no solution?

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.

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