# Course 1, Week 2, Practice Quiz, Question 6

The question:
In the following matrix:
[ a a ]
[ b c ]
a, b, and c are non-zero real numbers. If the matrix is non-singular, which of the following must be true:

1. c = a only if a = b
2. a = b only if c ≠ a
3. c ≠ b
4. c = b

I’ve checked answer 3: c ≠ b
and I got a feedback: “You didn’t select all the correct answers”, but i don’t understand how the other answers can be correct:

1. if a = c and a = b => the matrix will be:
[ a a ]
[ a a ]
this is a singular matrix (redundant) - so this answer is false.

2. if a = b only if c ≠ a => the matrix will be:
[ a a ]
[ a c ]
this is also a singular matrix (contradictive) - so this answer is false.

3. if c = b => the matrix will be:
[ a a ]
[ b b ]
and i can say that aZ = b (Z is some constant) so:
[ a a ]
[ aZ aZ ]
and therefore, the rows are dependent => this is a singular matrix (redundant)

so i clearly missing something here… which of these answers is correct, and why?

1 Like

@Omri_Suissa

In case of 2, the determinant is
ac - a^2 \neq 0
[a \neq c]

Try “the determinant” in W1

Thank you!
I’ve been looking at it incorrectly; I can not assume that the matrix is contradictory since I don’t know the y (the result of the equation).

1 Like

Hey @AeryB,
Don’t you think that selecting the option a = b only if c != a would be incorrect, as per the wording of the question? The question is as follows:

a, b, and c are non-zero real numbers. If the matrix is non-singular, which of the following must be true:

The word to note here is “must”. For non-zero determinant (non-singular matrix), a * (c-b) != 0, i.e., the only conditions that must be true are a != 0 and c != b. The second condition could be true, but it’s not a condition that must be true for the matrix to be non-singular. What are your thoughts on this?

Cheers,
Elemento

Hello @Elemento,

Your doubt sure got me confused too!

Well, watch out!
Both of the following forms are equivalent:

i. a = b only if c ≠ a
ii. if a = b, then c ≠ a

Source: 1, 2

Finally, say,

1. ‘the matrix is non-singular’ \equiv p
2. q \in \{1, 2, 3, 4\} (set of options)

When we consider p to be T (true) and q is inferred T from it, we add q to the set of correct option(s). [Our statement is p \to qreference here! Especially look at the first table!]

If you followed the reference,
you’d observe that if our p results 1 to be T, we are fine with it.
Similarly, if 2 results in T when p is T, we’ll add it to our set of correct options.

I know it may look really confusing (), but try to understand why it makes a contradiction with your statement:

Best wishes…^^

Hey @AeryB,
Thanks a lot for the detailed explanation. As you said, it’s a bit confusing indeed.

If both are equivalent (which they are), don’t you think, that it would be easier for the learners to interpret the second statement than the first statement? I guess it would make the option much more intuitive, if we replace the existing statement with it’s other variation as you have mentioned.

Cheers,
Elemento