# Another true / false question

#### Yankel

##### Active member
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 ?

MHB Math Helper

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

##### Well-known member
MHB Math Scholar
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 )