Below is a plot of land divided into 4 sections for a vegetable garden.
1 answer:
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Answer:
C.
Step-by-step explanation:
Hi there!
To answer this, we must set it up as
Now we just use the distributive property to solve.
this becomes
I hope this helps!
A. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).
b. True. Suppose
is the matrix of the eigenvectors of 
, and 
is the diagonal matrix of the eigenvalues of :
Then
In other words, the columns of
are 
, which are identically 
, and these are the columns of 
.
c. False. A counterexample is the matrix
which is nonsingular, but it has only one eigenvalue.
d. False. Consider the matrix
with eigenvalue
and eigenvector 
, where 
. But the matrix can't be diagonalized.
b. True. Suppose
Then
In other words, the columns of
c. False. A counterexample is the matrix
which is nonsingular, but it has only one eigenvalue.
d. False. Consider the matrix
with eigenvalue