How do you solve #2x+3y=-1# and #4x-5y=7# using matrices?

Although the method might seem quite daunting, once the preparation process is mastered, the method itself is surprisingly quick and easy, involving a few simple calculations.

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We have the following equations:

#2x+3y =-1 " and "4x-5y =7#

First write them as matrices:

#((2," "3),(4,-5)) ((x),(y)) = ((-1),(7))#

Now find the inverse matrix of #A =((2," "3),(4,-5))#

#abs(A) = (2xx -5)-(4xx3) = -10-12 = -22#

#A^-1 = 1/-22((-5,-3),(-4,2)) = (color(red)((5/22,3/22),(2/11,-1/11)))#
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Multiply both sides of the matrix equation by the inverse matrix.

#(color(red)((5/22,3/22),(2/11,-1/11))) ((2," "3),(4,-5)) ((x),(y)) = (color(red)((5/22,3/22),(2/11,-1/11)))((-1),(7))#

#color(white)(xxxxxxxxx)((1,0),(0,1)) ((x),(y)) = ((-1/3),(1/2))#

#color(white)(xxxxxxxxxxxxxx)((x),(y)) = ((8/11),(-9/11))#

# :. x =8/11 and y = -9/11#

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Background knowledge... to help with the method above..
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A 2 x 2 matrix multiplied by the unit matrix remains unchanged

# ((1,0),(0,1)) ((a,b),(c,d)) = ((a,b),(c,d))#

A matrix multiplied by its inverse gives the unit matrix -
also known as the Identity Matrix.

# A xx A^-1 = I = ((1,0),(0,1))#

To find the inverse matrix (#M^-1#) of matrix M

#M = ((a,b),(c,d))#

1. Find the determinant #(abs(M)) = ad-bc#

2. #M^-1 = 1/((abs(M)))( ( d,-b),(-c,a))#

(swop a and d and change the signs of b and c, then divide by the determinant.)