3[4 x]+ 2[y -3 ] = [10 0] output for the matrices
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Step-by-step explanation:
Taking the L.H.S, we have
3[4 x]+2[y −3]=[12 3x]+[2y −6]=[(12+2y) (3x−6)]
Now, equating with R.H.S we get
[(12+2y) (3x−6)]=[10 0]
12+2y=10 and 3x−6=0
2y=−2 and 3x=6
y=−1 and x=2
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