Is Statement 3 Correct Concerning Non-Invertible Matrices in Linear Systems?

[Yankel]

Hello,

I got one more true / false question, this time I need to pick the CORRECT answer:

1. If Y is a non trivial solution of a system Ax=b, then 2Y is also a solution of this system

2. If a system (Ax=b) has 4 equations, 6 unknowns and infinite number of solutions with 2 degrees of freedom, than the rank of A is 2.

3. Let A be a 3X3 matrix so that the system Ax=0 has a solution x=(1 0 3)^t, then A is not invertible.

4. The system Ax=b with m equations and n unknowns (m>n) can not have a single solution.

5. All the statements are wrong

I think that 2 and 4 are wrong. 1 sounds weird, I do not know of trivial / non trivial solution for Ax=b...only for Ax=0...
that leaves me with 3, I am not sure about it, is it correct that if the solution exist, than the system has infinite number of solutions and not just the trivial one ? and in this case, if A is invertible then:

A^-1Ax=A^-10 --> x=0 ?

so 3 is the correct one ?

 

[Fernando Revilla]

so 3 is the correct one ?

Right.

 

[I like Serena]

Hello,

I got one more true / false question, this time I need to pick the CORRECT answer:

1. If Y is a non trivial solution of a system Ax=b, then 2Y is also a solution of this system

1 sounds weird, I do not know of trivial / non trivial solution for Ax=b...only for Ax=0...

This does sound weird, since the symbol Y has not been defined.
But from the context I deduce that they mean that Y is supposed to be a solution for x.

A trivial solution would be x=0, which would be a solution if b=0.
So we assume that Y is not equal to 0.

So suppose AY=b, what do you think A(2Y) is?

2. If a system (Ax=b) has 4 equations, 6 unknowns and infinite number of solutions with 2 degrees of freedom, than the rank of A is 2.

How many rows and columns do you think A has?
How many solutions would you expect (how many degrees of freedom) if A is a normal, non-degenerate matrix?
What would be the rank of A in that case?

 

[Deveno]

the way you show a statement is false, is you exhibit a counter-example. it need not be fancy, and ANY counter-example will do.

so let's falsify statement 1 with a particular A, y and b that makes it untrue. here is my choice:

$A = \begin{bmatrix}1&0\\0&0 \end{bmatrix};\ y = \begin{bmatrix}1\\1 \end{bmatrix};\ b = \begin{bmatrix}1\\0 \end{bmatrix}$

then Ay = b, but A(2y) = A(2,2) = (2,0) ≠ b.

again, we can falsify 2 in a similar manner. we just need to find a system of 4 equations in 6 unknowns with 2 degrees of freedom, and show that the matrix A for this system does not have rank 2. how about this system:

$\begin{array}{c}x_1 + x_5 = 2\\x_2+x_6 = 2\\x_3 = 1\\x_4 = 1\end{array}$

it has (augmented) matrix:

$\left[\begin{array} {cccccc|c}1&0&0&0&1&0&2\\0&1&0&0&0&1&2\\0&0&1&0&0&0&1\\0&0&0&1&0&0&1 \end{array}\right]$

does this matrix have rank 2?

again, we can falsify 4. we are free to pick any m,n with m > n that suits us. let's pick m = 2, n = 1. so we have 2 equations in one unknown. how about these?

x = 1
2x = 2

this system has the solution x = 1. here, our matrix A is:

$\begin{bmatrix}1\\2\end{bmatrix}$

and $b = \begin{bmatrix}1\\2 \end{bmatrix}$.

so we're down to #3 and #5. clearly only one of these can be true. let's be fancy, and prove #3 by contradiction.

suppose (1,0,3) is in the nullspace of an invertible matrix A:

A(1,0,3) = (0,0,0).

applying A^-1 to both sides, we get:

(1,0,3) = A^-1(0,0,0) = (0,0,0), which is absurd. so no such matrix can exist.

so if we have a matrix A with A(1,0,3) = (0,0,0), then A cannot be invertible. this proves #3 and thus disproves #5 (it is a counter-example :P)

 

[blue_raver22]

I can provide the following response to the content provided:

1. True. If Y is a non-trivial solution of Ax=b, then 2Y is also a solution because multiplying both sides of the equation by a non-zero constant does not change the solution set.

2. False. The rank of A in this case would be 4, not 2. The number of degrees of freedom does not determine the rank of A.

3. True. If the system Ax=0 has a non-trivial solution, then A is not invertible. In this case, x=(1 0 3)^t is a non-trivial solution, so A would not be invertible.

4. False. The system Ax=b with m equations and n unknowns (m>n) can have a single solution if the system is consistent (i.e. there is a solution for b). However, if m>n and the system is inconsistent, then there would not be a single solution.

5. False. Only statements 2 and 4 are incorrect. Statements 1, 3, and 5 are correct.

In summary, the correct answer is 5, as not all statements are wrong. It is important to carefully consider the definitions and properties of solutions and matrices when determining the correctness of statements in linear algebra.