What is going to be the easiest approach?

[mr_coffee]

Hello everyone, I was wondering what the easiest approach would be to do this:
Find values of k for which the matrix system Ax = b, has a) no solutions, b) infintie solutions, and c) a unique solution:
A =
1 1 2
1 2 -3
7 17 k

b =
1
-3
32

I remeber doing a homework problem that said, Find the value of k for which the matrix A had rank 2, it also had a k in it, but it was a 3x3 matrix, this one has a vector b, so i don't know what to do with that, or if i can ignore it. With my old homework problem, I found the value of k which would make the matrix A rank 2, by finding the determinant of the matrix and setting it equal to 0, then solving for k and it worked! But really what did i find? Did i find a case in which the system had no solutions by doing that? because if its rank 2, and its a 3x3, then it should be a rank 3 if it had a unqiue solution right? Sorry I'm getting off the subject of my orngial question but i think it might help me solve this also. But my real question i need help with is my first question above! thanks!

 

[benorin]

Hint: set det(A)=0, and solve for k.

 

[mr_coffee]

Thanks, but this will solve for all cases? I thought it would just solve for the no solutions

 

[amcavoy]

Actually, that will tell you which values of k will give infinitely many solutions as well. A property of determinants is that if two rows are dependent, the determinant is zero. Now, if there is a zero row, the determinant will also be zero. A unique solution will occur when you can row-reduce and get the last row to look like: [0 0 f(k)] where f(k) is non-zero. I would recommend row-reducing here to get it into echelon form.

 

[HallsofIvy]

I would think row-reducing to an upper triangular matrix would be simplest. Any value of K which makes the last entry in the last row non-zero will give a unique answer. Any value of k which makes that last entry zero will give
a) an infinite number of answers if the last term in the reduced AUGMENTED matrix is also 0
b) no answer if it is not 0.

 

[mr_coffee]

Ohhh i c! Our professor showed us this way, and he would like us to do it this way on the exam. But I'm alittle confused on the very last part, but I'll show you this because you will probably be like wtf is he talking about. He expanded along the first row:
det A = 1(2k+51)-1(k+21)+2(17-14) = k + 36;
k = -36, so as long has k != 36 we got 1 unqiue solution.

THen finding the next part was kinda tricky, and I'm lost on.
A =
1 1 2 1
1 2 -3 -3
7 17 -36 32

so you know
a(1 1 2) + b(1 2 -3) = 7 17 -36 if there is infinite many solutions. <-- i get this part.
a+b = 7;
a + 2b = 17;
2a -3b = -36;
so u can see from this that a = -3, and b = 10;

Okay now this is where I'm lost, he wrote down the following:
-3(1) + 10(-3) = -33;
-33 != 32, so there is no solutions, and there is no case when k can produce infintatly many. I understand if u got 32 = 32, then there would be inftinatley many solutions but where did he get -3(1) + 10(-3)? It looks like he is plugging in 1 and -3, but into wht exactly? THanks for that method as well ivey.