Solution:
(5.1):
A=[131−1]∣A∣=∣∣131−1∣∣=1(−1)−3(1)=−4 ...(i)
−2A=−2[131−1]=[−2−6−22]∣−2A∣=∣∣−2−6−22∣∣=−2(2)−(−2)(−6)=−4−12=−16
Now, ∣−2A∣=−16=−4(4)=4A [Using (i)]
(5.2):
A=⎣⎡−212143351⎦⎤∣A∣=∣∣−212143351∣∣=−2⋅det(4351)−1⋅det(1251)+3⋅det(1243)
=−2(−11)−1⋅(−9)+3(−5)=16 ...(i)
Now, −2A=−2⎣⎡−212143351⎦⎤
=⎣⎡4−2−4−2−8−6−6−10−2⎦⎤
∣−2A∣=4⋅det(−8−6−10−2)−(−2)det(−2−4−10−2)−6⋅det(−2−4−8−6)
=4(−44)−(−2)(−36)−6(−20)=−128=16(−8)=−8∣A∣ [using (i)]
Comments