Help needed with previous exam questions

[mrb3n]

Can someone help me out here. I've tried to answer these questions from a few different angles but to no success.

[saba.ay]

With Question 8, have you tried subbing in the given values for k and seeing which value of k results in the inverse being the same as the original matrix? There's another mathematical/better approach to this but I don't know how to use latex.

Also with Question 9, It's very much just trial and error, but be systematic in your approach. I'd do this by entering Matrix B and making all the numbers 1 for convenience and entering Matrix A with all values 0, except two. Again, I'd number these 1 as well for convenience. I'd see the number of 0 I have when I multiply these two matrices. I'd then repeat the process, but change the location of the two which aren't numbered 0 in Matrix A. I'd move them to another row/ column, etc. Again, with this question, there is somewhat of a more mathematical approach which I remember writing on my cheat sheet on the exam last year, but I've thrown that out, so... sorry?

[KevinooBz]

Can someone help me out here. I've tried to answer these questions from a few different angles but to no success.

Q8) First, to find the inverse you need the determinant
det(A)= (-3 x 3) - (-4 x k) = 4k-9
A^-1= 1/4k-9 times A, when you have your inverse matrix you can just equate any of the elements and solve for k.
-3/(4k-9)=3
-3=12k-27
24=12k
D)k=2
Q9) This is just testing your knowledge. Matrix A is a 3x3 matrix. If the product matrix AB is defined, B has to be a 3x2 matrix. If 7 elements from A are 0, the last two have to non zero. From the information, to have a product of zero in the matrix AB, any row or column in matrix A has to have 3 consecutive elements of 0. If you have two non zeros next to each other in matrix A, four elements in matrix AB will be zero. If you have two non zeros that are not next to each other in matrix A, either two or three elements in AB will be zero. It is impossible to have 6 zeros in the matrix AB and 2 is already greater than 0 and 1.
Therefore, the minimum is C)2.
You can test this out for yourself using 0's and any random variable you choose that is not 0, I used x's.

[mrb3n]

Thanks for the help guys