Question 355472
<pre><font size = 4 color = "indigo"><b>

{{{system(3x-4y=-1,  
-2x+3y=3)}}}

Write that system in the matrix form AX = B


{{{(matrix(2,2,3,-4,-2,3))(matrix(2,1,x,y))=(matrix(2,1,-1,3))}}} 

First find the inverse of {{{(matrix(2,2,3,-4,-2,3))}}}:

To do that:

1.  Find the value of its determinant, {{{abs(matrix(2,2,3,-4,-2,3))=(3)(3)-(-4)(-2) = 9-8 = 1}}}

2. Swap the upper left and lower right elements of {{{(matrix(2,2,3,-4,-2,3))}}},
getting {{{(matrix(2,2,3,-4,-2,3))}}}

[In this case it didn't change anything since they were both 3, but in
other problems it will be different and you must swap them]

3. Then change the signs of the upper right and lower left elements,
getting {{{(matrix(2,2,3,4,2,3))}}}

4. Divide every element by the value of the determinant of the 
original matrix which we found to be 13 in step 1, getting
{{{(matrix(2,2,3/1,4/1,2/1,3/1))}}} or {{{(matrix(2,2,3,4,2,3))}}}

[In this case it didn't change anything since the determinant was 1, but
in other problems it will be different and you must divide.]

This is the inverse A<sup>-1</sup> of
the original matrix A.

Left-multiply both sides of the given matrix equation, AX=B

{{{(matrix(2,2,3,-4,-2,3))(matrix(2,1,x,y))=(matrix(2,1,-1,3))}}}

by this inverse, getting the form A<sup>-1</sup>(AX)=A<sup>-1</sup>B

{{{
(matrix(2,2,3,4,2,3))((matrix(2,2,3,-4,-2,3))(matrix(2,1,x,y)))=

(matrix(2,2,3,4,2,3))(matrix(2,1,-1,3))}}} 

Use the associative principle to move the parentheses around
the first two matrices on the left, getting the form (A<sup>-1</sup>A)X=A<sup>-1</sup>B

{{{
((matrix(2,2,3,4,2,3))(matrix(2,2,3,-4,-2,3)))(matrix(2,1,x,y))=

(matrix(2,2,3,4,2,3))(matrix(2,1,-1,-3))}}} 

Do the matrix multiplication:

{{{(matrix(2,2,(3)(3)+(4)(-2), (3)(-4)+(4)(3),
(2)(3)+(3)(-2), (2)(-4)+(3)(3) ))(matrix(2,1,x,y))=
(matrix(2,1,(3)(-1)+(4)(3), (2)(-1)+(3)(3))) }}}

Simplify:

{{{ (matrix(2,2,9-8, -12+12, 6-6, -8+9 ))(matrix(2,1,m,n))=(matrix(2,1, -3+12,  -2+9)) }}}

Simplify some more:

{{{ (matrix(2,2,1, 0, 0, 1 ))(matrix(2,1,x,y))=(matrix(2,1, 9, 7)) }}}

This is the form IX=A<sup>-1</sup>B


Multiply the matrices on the left:

{{{(matrix(2,1,1*x+0*y,0*x+1*y))=(matrix(2,1,9,7))}}}

Simplify:

{{{(matrix(2,1,x,y))=(matrix(2,1,9,7))}}}.  This is the form X=A<sup>-1</sup>B.

So the solution is {{{x=9}}} and {{{y=7}}}

Edwin</pre>