Given system of equations5x+2y=3
3x+2y=5
This can be written as
AX=B
where A=[5232],X=[xy],B=[35]
Here, |A|=10−6=4
Since, |A|≠0
Hence, A−1 exists and the system has a unique solution given by X=A−1B
A−1=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 [2−3−25]
⇒adjA=CT=[2−2−35]
⇒A−1=adjA|A|=14[2−2−35]
The solution is X=A−1B
[xy]=14[2−2−35][35]
[xy]=14[6−10−9+25]
⇒[xy]=[−14]
Hence, x=−1,y=4