Two Further Maths Questions

1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?

1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?

I think this is correct?? Let me know.
It’s just playing around with the matrix identities.
So essentially M^-1 is equal to the transposed Matrix

Hence why only 1 Mark. I have absolutely no idea if that correct though...

I think this is correct?? Let me know.
It’s just playing around with the matrix identities.
So essentially M^-1 is equal to the transposed Matrix

Hence why only 1 Mark. I have absolutely no idea if that correct though...

1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?

9c. You need to get it in the form x+iy before you can isolate the real part like that. So, mupltiply top & bottom by the complex conjugate of the denominator.

9c. You need to get it in the form x+iy before you can isolate the real part like that. So, mupltiply top & bottom by the complex conjugate of the denominator.

You CAN cancel IDENTITY matrices...
Hence why I cancelled them out. My answer of M^T = K(M^-1) is what you achieved also...

For 9d I did this but got sin 2x = -4 which
is obviously invalid so can you see what I did wrong?

That's what I effectively said, but more importantly I told you to not get into the habit of cancelling out matrices between both sides. You cannot divide by a matrix and multiplication is not commutative.

$ABC = DBE$ does not mean $AC = DE$.

You should take these properties more seriously by instead of putting a slash through $I$, instead note that $I$ is such a matrix so that $AI=IA=A$. Also, since we don't have commutativity, you need to pre-multiply or post-multiply by $M^{-1}$. Hence the correct practice to get into for matrices is as follows;

$MM^T = kI$

$M^{-1}MM^T = kM^{-1}I$

$IM^T = kM^{-1}$

$M^T = kM^{-1}$


where at no point have I 'cancelled' out common matrices between both sides, and I have strictly pre-multiplied both sides by $M^{-1}$.

I would suggest checking the formula you have for S - you've not posted its derivation, but I'd guess that's where the problem lies.

Can you see what I need to do for 9d?

Don't have time to look into this too much at the moment, but for 8d have you tried solving the following system?

$q+4r-s=0$

$3q + ps = 0$

$q^2 + r^2 + s^2 = 18$

I had set up those three but the squares put me off... I will try again! Thanks

1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?

1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?