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.)