Question 732098
<pre><font size = 4 color = "indigo"><b>
Here is one exactly like it.  Use it as a guide: 

{{{system(x+2y=1,4x+5y=13)}}}

Abbreviate the above system by the AX=B form:

{{{(matrix(2,2,1,2,4,5))(matrix(2,1,x,y))=(matrix(2,1,1,13))}}}

Now we need to find the inverse of the coefficient matrix

{{{(matrix(2,2,1,2,4,5))}}}

To find the inverse of a 2x2 matrix:

1. Interchange the upper left and lower right elements:

{{{(matrix(2,2,5,2,4,1))}}}

2. Multiply the upper right and lower left elements by -1:

{{{(matrix(2,2,5,-2,-4,1))}}}

3. Find the determinant of this matrix:

{{{abs(matrix(2,2,5,-2,-4,1))=(5)(1)-(-2)(-4)=5-8=-3}}}

4. Divide every element of {{{(matrix(2,2,5,-2,-4,1))}}} by this value:

{{{(matrix(2,2,5/(-3),(-2)/(-3),(-4)/(-3),1/(-3)))}}}

5. Simplify

{{{(matrix(2,2,-5/3,2/3,4/3,-1/3))}}}

That is the inverse of the coefficient matrix.

Left-multiply both sides of the matrix
equation:

{{{(matrix(2,2,1,2,4,5))(matrix(2,1,x,y))=(matrix(2,1,1,13))}}}

by the inverse of the coefficient matrix:

{{{

(matrix(2,2,-5/3,2/3,4/3,-1/3))

((matrix(2,2,1,2,4,5))(matrix(2,1,x,y)))=


(matrix(2,2,-5/3,2/3,4/3,-1/3))
(matrix(2,1,1,13))}}}

Since matrix multiplication is associative, we move
the parentheses:

{{{

((matrix(2,2,-5/3,2/3,4/3,-1/3))

(matrix(2,2,1,2,4,5)))(matrix(2,1,x,y))=


(matrix(2,2,-5/3,2/3,4/3,-1/3))
(matrix(2,1,1,13))}}}

Now we multiply the two matrices on the far
left and the far right:

{{{

( matrix(2,2,

(-5/3)(1)+(2/3)(4),(-5/3)(2)+(2/3)(5),(4/3)(1)+(-1/3)(4),(4/3)(2)+(-1/3)(5)))

(matrix(2,1,x,y))=(matrix(2,1,(-5/3)(1)+(2/3)(13),(4/3)(1)+(-1/3)(13)))}}}

Simplify:

{{{

( matrix(2,2,

-5/3+8/3,-10/3+10/3,4/3-4/3,8/3-5/3))

(matrix(2,1,x,y))=(matrix(2,1,-5/3+26/3,4/3-13/3))

}}}

{{{

( matrix(2,2,

3/3,0/3,0/3,3/3))

(matrix(2,1,x,y))=(matrix(2,1,21/3,-9/3))

}}}

{{{( matrix(2,2,

1,0,0,1))

(matrix(2,1,x,y))=(matrix(2,1,7,-3))

}}}

Multiply the matrices on the left:

{{{(matrix(2,1,(1)(x)+(0)(y), (0)(x)+(1)(y)))=(matrix(2,1,7,-3))}}} 

Simplify:

{{{(matrix(2,1,x,y))=(matrix(2,1,7,-3))}}} 

So the solution is 

{{{x=7}}}, {{{y=-3}}}

Edwin</pre>