Linear Algebra_25

Question

Let AX=B be a system of n linear equations in n unknown with integer coefficient and the components of B are all integer. Consider the following

(1)det(A)=1

(2)det(A)=0

(3)Solution X has integer entries

(4)Solution X does not have all integer entries

For the given system of linear equations which one of the following is correct?

(a)Only 3 unconditionally holds true.

(b)If 1, then 3 holds true

(c)If 1, then 4 holds true

(d)If 2, then 3 holds true

I think (d) should be the answer.

Comments

@ Deepanshu

why C?

just take the equation x-y=0,y=1. solving it, we will get solutions as integer. 1,1

we can also take other counter examples.

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aambazinga whats problem with c ?????

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@srestha-A is the coefficient matrix. X is the solution vector

Answer 1

By Cramer's rule, we know that element $x_i$ of solution matrix $X$ can found by the following formula:

$x_i=\frac{det(A_i)}{det(A)}$ 

assuming $det(A)=1$ , we get $x_i=det(A_i)$    

(recall when applying cramer's rule, $A_i$ is a matrix formed by replacing the $i^{th}$ column in matrix $A$ with column matrix $B$)

It is already given that elements of $B$ are integral and all coefficients in equations are integral therefore elements of matrix $A$ are integral as well. As $A_i$ is made from elements of $A$ and $B$, therefore it's elements are integral as well.

Determinant of a matrix with integral elements will always be integral (you can prove this yourself) therefore $det(A_i)$ will also be integral. As $x_i= det(A_i)$ therefore $x_i$ is integral as well.

Hence we have shown that all elements of solution matrix  X will be integral when $det(A)=1$. 

So answer is (b)

 

Comments

bro whts problem with c?????

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What you are stating is that if det(A) =1 then it does not have all integral solutions, which is exactly opposite of what is proved above. @aambazinga has already shown you a counter example in his comment.

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I think it can be (d) also.

if det(A)=0, means the matrix is singular.

Means the rows and columns of A are dependent and hence I can design a non-zero vector X such that AX=O the solution to the null space and if B contains all zeroes which are integer, AX=O holds.

Answer 2

Can it not be answered using the following logic.

When det(A) = 0, it means that our vectors are linearly dependent and therefore the set of solutions will be dependent on the free variable, so if the free variable is suppose ‘z’ then the solution set will be in terms of ‘z’, hence non-integral.

or it can be that the system is inconsistent and no solution exists for x.

Whereas when det(A) != 0, it means that the vectors are linearly independent and hence unique solution exists and therefore x will have all integral values.

Please correct me if wrong.