1. The standard matrix of the linear operator R4 -> R4 defined by the formula
T(a,b,c,d) = (4a-3b+4c-d,3a-5b+d,a+d,b+d)
is
| A := |
[ 4 |
-3 |
4 |
-1 ] |
| [ 3 |
-5 |
0 |
1 ] |
| [ 1 |
0 |
0 |
1 ] |
| [ 0 |
1 |
0 |
1 ] |
To test this, we can define the vector v and check if A*v = b.
| v := |
[ a ] |
| [ b ] |
| [ c ] |
| [ d ] |
> b:= evalm(A&*v);
[ 4 a - 3 b + 4 c - d ]
[ 3 a - 5 b + d ]
v := [ a + d ]
[ b + d ]
It works. Our standard matrix, A, of the linear operator T was correct.
2. Is there a solution of the equation T(v) = (1,1,1,1) where T is the operator defined in Problem 1?
We have the system of equations
4a - 3b + 4c - d = 1
3a - 5b + d = 1
a + d = 1
b + d = 1.
This system has the augmented matrix
| A := |
[ 4 |
-3 |
4 |
-1 |
1 ] |
| [ 3 |
-5 |
0 |
1 |
1 ] |
| [ 1 |
0 |
0 |
1 |
1 ] |
| [ 0 |
1 |
0 |
1 |
1 ] |
The reduced row-echelon form of this augmented matrix is
| B := |
[ 1 |
0 |
0 |
0 |
0 ] |
| [ 0 |
1 |
0 |
0 |
0 ] |
| [ 0 |
0 |
1 |
0 |
1/2 ] |
| [ 0 |
0 |
0 |
1 |
1 ] |
So, v = (a,b,c,d) = (0,0,1/2,1).
3. Find the standard matrix for the reflection of R2 about the axis with direction (1,2).
The axis with direction (1,2) is y = 2x. We know, if we draw the line between a point and its reflection it must be perpendicular to y = 2x. So, it has slope -1/2. We also know a point is an equal distance from the line y = 2x as its reflection.
If we pick a point, say (2,0), then the line containing the point and its reflection is y = (-1/2)x + 1. The point of intersection between this line and y = 2x is (2/5,4/5). The reflection point is an equal distance from the intersection point as (2,0), hence the reflection point is (-6/5,8/5). If we do this for other points we find
(1,0) -> (-3/5, 4/5)
(2,0) -> (-6/5, 8/5)
(1,1) -> (3/5, 6/5).
We know if the original point is the vector v and the reflected point is the vector b, then A*v = b. So, if we say
If we plug in the point (1,0) and its reflection point (-3/5,4/5) into the equation A*v = b, we find a = -3/5 and c = 4/5. Then we plug in the point (1,1) and its reflection point (3/5,6/5) to find b = 6/5 and d = 2/5. So, the standard matrix is
| A := |
[ -3/5 |
6/5 ] |
| [ 4/5 |
2/5 ] |