Section 4-4

 

(2)

[
-3
5
]
2
0
1
4
+
[
2
1
]
-6
3
0
-5
=
[
-1
6
]
-4
3
1
-1

(8)

10[
2
-1
3
]
0
-4
5
=
[
20
-10
30
]
0
-40
50

(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.)