The linear transformation illustrate that:
a. There are infinite solutions.
b. 1
c. 2
d. 3
Since A is the matrix of the linear transformation T, hence ker(T) = ker(A), nullity (T) = nullity(A), range(T) = col(A) and rank(T) = rank(A).
(a) ker(T) = ker(A) is the set of solutions to the equation T(X) = AX = 0. To solve this equation, we will reduce A to its RREF as under:
Interchange the 1st row and the 2nd row
Multiply the 1st row by 1/12
Multiply the 2nd row by -1/6
Then the RREF of A is
1 0 19/12
0 1 -7/6
Now, if X = (x,y,z)T, then the equation T(X) = AX = 0 is equivalent to x +19z/12= 0 or, x = -19z/12 and y-7z/6 = 0 or, y = 7z/6 so that X = (-19z/12,7z/6,z)T = (z/12)(-19,14,12)T.
Thus, ker(T) = {t(-19,14,12)T} where t is an arbitrary real number. There are infinite solutions.
(b). Since the RREF of A has 2 non-zero rows, hence rank(A) = 2. As per the rank-nullity theorem, nullity(T) = nullity(A) = No. of columns in A, -rank(A) = 3-2= 1.
( c). The first 2 columns of A are linearly independent and its 3rd column is a linear combination of its first 2 columns.
Hence, range(T) = col(A) = span{(0,12)T,(-6,0)T} = R2 ( as the vectors (0,12)T,(-6,0)T are scalar multiples of (0,1)T,(1,0)T respectively and the vectors (0,1)T,(1,0)T constitute the standard basis of R2.
(d). As in (b) above, rank(T) = rank(A) = 2.
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