If A=220211−72−3 and if A3−13A+kI=0, then k=.
2
-2
-12
12
We have A3−β1A2+β2A−β3I=0
Where β1=0;β2=−13;β3=−12
∴A3−13A−(−12)I=0⇒A3−13A+12I=0
Clearlv k=12