PLEASE HELPPP! I'LL GIVE BRAINLEST
1 answer:
You might be interested in
By inspection, it's clear that the sequence must converge to 
because
when
is arbitrarily large.
Now, for the limit as
to be equal to is to say that for any 
, there exists some 
such that whenever 
, it follows that
From this inequality, we get
As we're considering, we can omit the first inequality.
We can then see that choosing
will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that 
.
when
Now, for the limit as
From this inequality, we get
As we're considering
We can then see that choosing
Answer:
U₂₁ and U₂₂ fits in the line of upper triangular
Step-by-step explanation:
As explain in the question
An upper triangular matric is square matrix with all zeros below its daignal. A lower triangular matrix is a square matrix with all zeros above its diagonal.
(as shown in the attached image)
(a)for all 3×3 upper triangular matirices U, verify that U2 is also upper triangular
for a matrices U (having a 3*3 matrix), assuming for example, U is given below:
U =
the element in the upper triangular are 1,2,3,5,6,9 (U₁₁, U₁₂, U₁₃; , U₂₂, U₂₃, U₂₂)
while lower triangular are 0,0,0, the element in the lower triangular are (U₂₁, U₃₁,U₃₂)
U₂₁ and U₂₂ fits in the line of upper triangular
