Question 274 of 472MediumNUMERICAL
Let A=100−1100−11 and B=7A20−20A7+2I, where I is the identity matrix of order 3×3. If B=[bij], then b13 is equal to ...
Answer: 910 Solutions · 1
Official solution
By induction, An=100−n10(2n)−n1.
b13=7(220)−20(27)+2(0)=7(190)−20(21)=1330−420=910. Answer the question to unlock