SOLUTION: {{{A=(matrix(2,2,2,-5,3,1))}}} Evaluate the square of A and find x,y,z, not all zero, such that the matrix xI+yA+zAA is the zero matrix.

Algebra ->  Matrices-and-determiminant -> SOLUTION: {{{A=(matrix(2,2,2,-5,3,1))}}} Evaluate the square of A and find x,y,z, not all zero, such that the matrix xI+yA+zAA is the zero matrix.      Log On


   



Question 341598: A=%28matrix%282%2C2%2C2%2C-5%2C3%2C1%29%29
Evaluate the square of A and find x,y,z, not all zero, such that the matrix

xI+yA+zAA is the zero matrix.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
A=%28matrix%282%2C2%2C2%2C-5%2C3%2C1%29%29
Evaluate the square of A and find x,y,z not all zero such that the matrix

xI+yA+zAA is the zero matrix.

We will find those three terms separately:

xI=x%28matrix%282%2C2%2C1%2C0%2C0%2C1%29%29=%28matrix%282%2C2%2Cx%2C0%2C0%2Cx%29%29

yA=y%28matrix%282%2C2%2C2%2C-5%2C3%2C1%29%29=%28matrix%282%2C2%2C2y%2C-5y%2C3y%2Cy%29%29

To find zAA we first find the square of A, which is AA



(That was the first thing asked for, the square of A)

Now we multiply that by z to get zAA:



Next we find their sum:



This must equal to the zero 2×2 matrix, so,



We form this system by setting each element = 0:

system%28x%2B2y-11z=0%2C-5y-15z=0%2C+3y%2B9z=0%2C+x%2By-14z=0%29

We solve that system by Gaussian elimination:



By using row operations we find the row reduced echelon form
of that augmented matrix.



The first two rows of that matrix gives us
the system:

system%28x-17z=0%2Cy%2B3z=0%29

or

system%28x=17z%2Cy=-3z%29

To find all solutions, we choose a constant k for z, then

system%28x=17k%2Cy=-3k%2Cz=k%29

Edwin