Multiplication of a Matrix by a scalar

Transcript

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