How do you solve the system $4x+8y=7, 3x-3y=0$ using matrix equation?

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Answer 1 · 2017-04-11T12:49:13

$x=7/12$ and $y=7/12$

In matrix notation, the system of linear equations

$4x+8y=7$ and $3x-3y=0$ can be written as

$((4,8),(3,-3))((x),(y))=((7),(0))$

i.e. $A((x),(y))=((7),(0))$ , then

$((x),(y))=A^(-1)((7),(0))$ ,

where $A^(-1)$ is inverse of $A=((a,b),(c,d))$ and for a 2X2 matrix is defined as $A^(-1)=1/(ad-bc)((d,-b),(-c,a))$

Hence inverse of $((4,8),(3,-3))$ is $1/(-12-24)((-3,-8),(-3,4))$

i.e. $1/(-36)((-3,-8),(-3,4))=((1/12,2/9),(1/12,-1/9))$

and $((x),(y))=((1/12,2/9),(1/12,-1/9))((7),(0))=(7/12,7/12)$

i.e. $x=7/12$ and $y=7/12$

How do you solve the system 4x+8y=7, 3x-3y=04x+8y=7, 3x-3y=0 using matrix equation?