Let A = [■8(3&2@1&4)]
3A = 3[■8(3&2@1&4)]
=[■8(3×3&3×2@3×1&3×4)]
= [■8(9&6@3&12)]
Let B = [■8(3&2@1&4@5&3)]
−5B = −5[■8(3&2@1&4@5&3)]
=[■8(−5×3&−5×2@−5×1&−5×4@−5×5&−5×3)]
=[■8(−15&−10@−5&−20@−25&−15)]
For matrix A = [■8(3&2@1&4)]
Negative of A = −A
= −1 × A
= −1 × [■8(3&2@1&4)]
= [■8(−3&−2@−1&−4)]
Similarly,
If X = [■8(−9&12&−8@5&6&0)]
−X = [■8(9&−12&8@−5&−6&0)]
Note: When calculating negative, we change the signs of all the elements

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.