Section 44
(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

3x4y

] = [

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 GaussJordan
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.)