18. e). A is a 5 x 4 matrix. Therefore, a = 4 and b = 5.
The RREF of A is
1 |
0 |
2 |
0 |
0 |
1 |
-1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
The kernel of T is the set of solutions to the equation T(X) = Ax = 0. In view of the RREF of A, if X = (x,y,z,w)T, then this equation is equivalent to x+2z = 0 or, x = -2z, y-z = 0 or, y = z and w = 0. Then X = (-2z,z,z,0)T = z(-2,1,1,0)T = t( -2,1,1,0)T where t = z is an arbitrary real number. This means that every solution to the equation T(X) = 0 is a scalar multiple of the vector ( -2,1,1,0)T. Hence the set {( -2,1,1,0)T } is a basis of the kernel of T.
The image of T is same as col(A). It may be observed from the RREF of A that the 1st, 2nd and 4th columns of A are linearly independent and the 3rd column is a linear combination of the first 2 columns.
Hence the set { ( 1,-1,0,3,1)T,( -2,-1,4,2,0)T, (0,0,3,-2,2)T} is a basis of the image of T.
f). The RREF of [A|b] is
1 |
0 |
2 |
0 |
0 |
0 |
1 |
-1 |
0 |
-2 |
0 |
0 |
0 |
1 |
-1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
This implies that b = 0a1-2a2-1a3+0a4. Hence b is in the span of the columns of A.
g). The vector ( 1,1,1,1,1)T is not in the column space of A as it cannot be expressed as a linear combination of the columns of A.
h). A is a 4 x 5 matrix . In view of the RREF of A, the system AX = b is a system of 3 linearly independent equations in 4 variables so that there will always be 1 free variable if the equation AX = b is consistent. Hence there cannot be a unique solution for any vector v
for which AX = v.
i). The set {a1,a2, a3, a4 } is not linearly independent as a3 = 2a1-a2.
j). It may be observed from the RREF of A that the set {a1,a2, a4 } is linearly independent as none of these 3 vectors may be expressed either as a a linear combination of the other 2 vectors or a scalar multiple of another vector.
k ).Since the set {a1,a2, a3, a4 } is not linearly independent, hence the set {a1,a2, a3, a4 , b} cannot be linearly independent.
Linear Algebra Question: 18. Consider the system of equations Ax = b where | A= 1...
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