T or F: If det(A^2-B^2)=0 and det(A+B) is not 0, then...

[mathstresser]

True or False? If det(A^2-B^2)=0 and det (A+B) is not equal to 0, then det(a)=det(B). Explain

False because one row or column from A^ can equal one row or column from B^, which would then make A^2-B^2 = 0, which would then make det(A^2-B^2)=0. This can occur when A does not equal B. So, it's very easy that det(A_B) is not equal to 0.

Is there anything I'm missing?

 

[daon]

Well the determinant mapping perserves multiplication, so:

det(A^2-B^2)=det(A+B)*det(A-B)

The determinant maps matricies to real numbers and since the real numbers have no zero divisors we have det(A+B)=0 or det(A-B)=0. We know the first is not the case, so det(A-B)=0.

Your implication is true then IFF det(A-B)=0 implies det(A)-det(B)=0. However we know the determinant does not perserve addition i.e. det(A+B) is NOT det(A)+det(B).

You needn't say this much, though, as to disprove something all you need to do is come up with a concrete counter example... no need to dig deep into theory.

 

[Count Iblis]

Well the determinant mapping perserves multiplication, so:

det(A^2-B^2)=det(A+B)*det(A-B)

The determinant maps matricies to real numbers and since the real numbers have no zero divisors we have det(A+B)=0 or det(A-B)=0. We know the first is not the case, so det(A-B)=0.

Your implication is true then IFF det(A-B)=0 implies det(A)-det(B)=0. However we know the determinant does not perserve addition i.e. det(A+B) is NOT det(A)+det(B).

You needn't say this much, though, as to disprove something all you need to do is come up with a concrete counter example... no need to dig deep into theory.

\(\displaystyle \left(A+B\right) \left(A-B\right)=A^{2}-B^{2}-AB+BA\neq A^{2}-B^{2}\)

 

[daon]

\(\displaystyle \left(A+B\right) \left(A-B\right)=A^{2}-B^{2}-AB+BA\neq A^{2}-B^{2}\)

Ah yes. My oops

 

[JakeD]

Well the determinant mapping perserves multiplication, so:

det(A^2-B^2)=det(A+B)*det(A-B)

The determinant maps matricies to real numbers and since the real numbers have no zero divisors we have det(A+B)=0 or det(A-B)=0. We know the first is not the case, so det(A-B)=0.

Your implication is true then IFF det(A-B)=0 implies det(A)-det(B)=0. However we know the determinant does not perserve addition i.e. det(A+B) is NOT det(A)+det(B).

You needn't say this much, though, as to disprove something all you need to do is come up with a concrete counter example... no need to dig deep into theory.

\(\displaystyle \left(A+B\right) \left(A-B\right)=A^{2}-B^{2}-AB+BA\neq A^{2}-B^{2}\)

I made the same mistake as daon. Thanks for pointing it out.

But what daon said is true for symmetric matrices, and good enough if we are just looking for a concrete counterexample:
\(\displaystyle \displaystyle A = \left[ \matrix{ 2 1 \\ 1 1} \right],\ B = \left[ \matrix{ 1 1 \\ 1 1} \right],\ A-B = \left[ \matrix{ 1 0 \\ 0 0} \right],\ \det{A-B} = 0 \text{ but }\det{A} = 1 \neq 0 = \det{B}.\)