Here is the problem:
Hello, I already tried the method shown by the OP of this post, "Find the values of a, b, c, and d from a matrix equation". Though, I'm not getting anywhere because the 2nd row and 2nd column of the variable matrix confuse me. I'm not sure what to do with the (d-a). First I applied scalar multiplication to the matrix on the right-hand side. Then, I did this:
- $a - b = 2$
- $ b - c = 2$
- $c - d = -6$
- $d - a = 2$
Once again, I already tried the method posted by the OP, which you can see in the link I've pasted. However, I'm still stuck and not sure what to do. Absolutely any help is appreciated, and thank you!
There's no unique answer. There's infinity answers here. The current style of the question may have confused you but you have to rewrite it as: A = [1 -1 0 0;0 1 -1 0;0 0 1 -1;-1 0 0 1] X =[a; b; c; d] and b = [2; 2; -6; 2] while AX=b. step 1) You may use these directly in MATLAB & you see that det(A) = 0 which means no unique answer. It means the number of answers are infinity like a line, a plane or a space. Here we see it's set of 3 linear equations. step 2) In MATLAB, we get the RREF fast (row reduced echelon form): rref(A) = 1
1 0 0 -1 0 1 0 -1 0 0 1 -1 0 0 0 0 remember that we had X = [a;b;c;d] so rref(A)X=g says we have these 3 lines at left side: a-d, b-d and c-d. Simply, I use your notes to create the right side of each of these 3 lines: a−b=2 b−c=2 c−d=−6 d−a=2
a-d = -2, b-d = -4 and c-d = -6. These 3 equation considered as answers while det = 0.