Good day to all. We have just completed our chapter on matrix operations (matrix

Question

Good day to all.

We have just completed our chapter on matrix operations (matrixmatrix product, special matrices etc...). In doing the end ofchapter exercises I have across a problem where the following isasked:

Question: Find all 2x2 matrices where A2 = 0

I considered the general 2x2 matrix A = [[a,b], [c,d]] where [a,b]and [c,d] are the rows of A.

Once I calculated AA I retrieved four equations which were thefollowing:

(1) a2 +bc = 0       =>bc = -(a2) => c= -(a2) / b with b <>0
(2) ab + bd = 0     => b(a+d) = 0 => b=0or a = -d
(3) ac + dc = 0     => c(a+d) = 0 => c=0or a = -d
(4) bc + d2 = 0   => bc = -(d^2)

From here I assumed that a = -d and constructed the followingmatrix:

A = [[a,b],[(-a2 / b), -a]] where the inner bracketsrepresent the rows of A.

I was just wondering if the process used is correct? I apologize ifthe presentation leaves a lot to be desired. I am not sure of howto code matrices. Any help would be greatly appreciated.

Kindest regards

Explanation / Answer

ok, let me have a stab at this [[a,b],[c,d]]^2 = 0 therefore (1) a2 +bc = 0      (2) ab + bd = 0     (3) ac + dc = 0    (4) bc + d2 = 0 4 eqn's, 4 unknowns, you should be able to solve this explicitly,sub 2 into 1 and 3 into 4 to obtain 5) (-bd/b)^2+bc=0, (-d)^2+bc=0 6) bc+(-ac/c)^2=0, bc+(-a)^2=0, therefore a=d now our 4 equations are (1) a2 +bc = 0      (2) ab + ba = 2ab= 0     (3) ac + ac = 2ac= 0    (4) bc + a2 = 0 , eliminate this one cuz its now thesame as 1 since 2ab=2ac, then b=c unless a and d=0 if a doesnt equal zero, then from eq 1, a2 + bc=0,since b=c, therefore both must be positive, hence a is definatelyequal to zero therefore a=d=0, reverting to the original equations, if a=d=0,then b and c must also be 0 therefore, a=b=c=d=0

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