यदि `A=[(1,-1),(2,3)]` तो दिखाइए कि `A^(2)-4A+5I=O`. इसकी सहायता से `A^(-1)` भी ज्ञात कीजिए।

Explanation

`A=[(1,-1),(2,3)]`
`impliesA^(2)=A.A`
`=[(1,-1),(2,3)][(1,-1),(2,3)]`
`=[(1-2,-1-3),(2+6,-2+9)]=[(-1,-4),(8,7)]`
अब L.H.S `=A^(2)-4A+5I`
`=[(-1,-4),(8,7)]-4[(1,-1),(2,3)]+5[(1,0),(0,1)]`
`=[(-1,-4),(8,7)]-[(4,-4),(8,12)]+[(5,0),(0,5)]`
`=[(0,0),(0,0)]=O=`R.H.S
अब `|A|=|(1,-1),(2,3)|`
`=3-(-2)=5!=0`
अत: `A^(-1)` का अ‌िस्‍तत्‍व है।
`:.A^(2)-4A+5I=0`
`impliesA^(-1)(A^(2)-4A+5I)=A^(-1).O`
`impliesA^(-1)(A^(2)-4A+5I)=A^(-1).O`
`impliesA-4I+5A^(-1)=0 :' A^(-1)A=I,A^(-1)I=A^(-1)`
`implies5A^(-1)=4I-A`
`=4[(1,0),(0,1)]-[(1,-1),(2,3)]`
`[(4,0),(0,4)]-[(1,-1),(2,3)]=[(3,1),(-2,1)]`
`impliesA^(-1)=1/5[(3,1),(-2,1)]`