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Question

Solve the system of equations, using matrix method
5x+2y=3,3x+2y=5

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Solution

Given system of equations
5x+2y=3
3x+2y=5
This can be written as
AX=B
where A=[5232],X=[xy],B=[35]

Here, |A|=106=4
Since, |A|0
Hence, A1 exists and the system has a unique solution given by X=A1B

A1=adjA|A| and adjA=CT

So, we will find the co-factors of each element of A.
C11=(1)1+12=2
C12=(1)1+23=3
C21=(1)2+12=2
C22=(1)2+25=5

So, the co-factor matrix is [2325]

adjA=CT=[2235]

A1=adjA|A|=14[2235]

The solution is X=A1B
[xy]=14[2235][35]

[xy]=14[6109+25]

[xy]=[14]
Hence, x=1,y=4

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