Find the matrix X such that XA=B ..?

#A = ((3a, 2b),(-a,b))#

Above is a single matrix A shown.

#B = (( -a , b ),(2a, 2b))#

Above shown is a single matrix B.

1 Answer
Truong-Son N.
May 29, 2017

#X = [(0,1),(4/5,2/5)]#

Wolfram Alpha confirms this.

Here is another way to do this. We write

#XA = B#,

and

#[(x_1,x_2),(x_3,x_4)][(3a,2b),(-a,b)] = [(-a,b),(2a,2b)]#.

From matrix multiplication, we obtain the following system of equations:

#3ax_1-ax_2 = a(3x_1 - x_2) = -a# #" "" "bb((1))#

#2bx_1 + bx_2 = b(2x_1 + x_2) = b# #" "" "" "bb((2))#

#3ax_3 - ax_4 = a(3x_3 - x_4) = 2a# #" "" "" "bb((3))#

#2bx_3 + bx_4 = b(2x_3 + x_4) = 2b# #" "" "" "bb((4))#

From #(1)#, we have:

#3x_1 - x_2 = -1#

#=> x_1 = x_2/3 - 1/3#

From plugging #x_1# into #(2)#, we have:

#2(x_2/3 - 1/3) + x_2 = 5/3 x_2 - 2/3 = 1#

#=> x_2 = 1#
#=> x_1 = 0#

From #(3)#, we have:

#3x_3 - x_4 = 2#

#=> x_3 = 2/3 + x_4/3#

From plugging #x_3# into #(4)#, we have:

#2(2/3 + x_4/3) + x_4 = 5/3x_4 + 4/3 = 2#

#=> x_4 = 2/5#
#=> x_3 = 12/15 = 4/5#

Therefore,

#color(blue)(X = [(0,1),(4/5,2/5)])#

#[(0,1),(4/5,2/5)][(3a,2b),(-a,b)] = [(-a,b),(2a,2b)]#.

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