2 For what value of k , kA kA 1, where A 1 . 1 t
Given :-
2 A = 1 1 ( kA) t ( kA) 1 To Find :k = ? Solution :First we will find (kA)
2 A = 1 1 2k k kA k ( kA) t
2 k
k -k
We know that (kA (k A) t (k (kA A)=1 )=1 Put the vaules
2k = 1 2k k -k k k 2 2 2 4k + k + k =1 6k 2 1 2
k
1
6 Taking square root on both sides k 2 k
1 6
1 6
So the value of k is
1 6
QUESTION No. 2 Determine the condition on , so that the Echelon and reduced Echelon form of
sin cos Given
cos
holds.
sin
:-
sin cos To Find =
cos
sin :-
?
Solution :-
sin cos
First we will make echelon form of
cos
by
sin
Divi Di vidi ding ng row row 1 wit with h sin sin , we wil willl get get
1 cos
cos / si s in
Add Cos Cos
sin
and th then di divide (s (sin +cos ) wi with ro row 2
cos / si s in 1 cos cos sin cos cos / si s in 1 0 (sin cos ) / ( sin cos ) 1 cot 0 1 ( sin / cos tan & 1/tan =cot ) The above
condition satisfies the echelon form
Now for Reduced echelon echelon form cot must be equal equal to zero cot = 0 =
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