Given, A3−5A2+7A+I=0 ⇒A5−5A4+7A3+A2=0 Also, A5−6A4+12A3−6A2+2A+2I =lA+mI A5−5A4+7A3+A2=0 Subtracting Eqs. (i) and (ii), we get ‌−A4+5A3−7A2+2A+2I=IA+mI ‌A4−5A3+7A2+A=0 On adding Eqs. (iii) and (iv), we get ‌3A+2I=lA+mI ⇒‌‌I=3‌ and ‌m=2 So, ‌‌l+m=5
