Section 4-4
(2)
|
+
|
|
=
|
|
(8)
|
=
|
|
(10)
|
[
|
-1
|
1
|
][
|
4
|
] |
=
|
[
|
(-1)(4)+(1)(2)
|
] |
=
|
[
|
-6 | ] |
|
2
|
-3
|
-2
|
(2)(4)+(-3)(-2)
|
14 |
(12)
| [ |
-3
|
2
|
][
|
-2
|
5
|
] = [
|
(-3)(-2)+2(-1)
|
(-3)5+2(3)
|
] = [
|
4
|
-9
|
] |
|
4
|
-1
|
-1
|
3
|
4(-2)+(-1)(-1)
|
4(5)+(-1)3
|
-7
|
17
|
(46)
| Find x and y so that |
[
|
5 | 3x |
] + [
|
1
|
-4y
|
] = [
|
6
|
-7
|
]. |
| 2x | -4 |
7y
|
4
|
5
|
0
|
| Adding matrices, this becomes |
[
|
6
|
3x-4y
|
] = [
|
1
|
-7
|
]. |
|
2x+7y
|
0
|
5
|
0
|
For the two matrices to be equal we must have x and y satisfying the following two equations:
| 3x |
-4y
|
=
|
-7
|
| 2x |
+7y
|
=
|
5
|
To find x and y we just need to solve the system. Of course we could use Gauss-Jordan
elimination, but that would be a bit of an overkill on such a simple system.
Let's use substitution instead. Solving for x in the top equation gives
x = 4/3 y - 7/3.
Plugging this into the second equation gives
2(4/3 y - 7/3) +7y = 5
8/3 y - 14/3 + 7y = 5
3( 8/3 y - 14/3 + 7y ) = 3(5)
8y - 14 +21y = 15
29y = 29
y=1
Now plugging y=1 back into the first equation gives us 3x - 4(1) = -7 which is 3x = -3, or x = -1.
Answer: Let x = -1 and y = 1 to make the statement true. (You can check
back to see that it really works.)