Here are the solutions to the matrix problems.
Given matrices:
A=[−1021]
B=[1−102]
a) Show that (A+B)(A+B)=A2+2AB+B2
Step 1: Calculate A+B.
A+B=[−1021]+[1−102]=[−1+10−12+01+2]=[0−123]
Step 2: Calculate (A+B)(A+B).
(A+B)(A+B)=[0−123][0−123]
=[(0)(0)+(2)(−1)(−1)(0)+(3)(−1)(0)(2)+(2)(3)(−1)(2)+(3)(3)]
=[−2−367]
Step 3: Calculate A2.
A2=A⋅A=[−1021][−1021]
=[(−1)(−1)+(2)(0)(0)(−1)+(1)(0)(−1)(2)+(2)(1)(0)(2)+(1)(1)]
=[1001]
Step 4: Calculate B2.
B2=B⋅B=[1−102][1−102]
=[(1)(1)+(0)(−1)(−1)(1)+(2)(−1)(1)(0)+(0)(2)(−1)(0)+(2)(2)]
=[1−304]
Step 5: Calculate AB.
AB=[−1021][1−102]
=[(−1)(1)+(2)(−1)(0)(1)+(1)(−1)(−1)(0)+(2)(2)(0)(0)+(1)(2)]
=[−3−142]
Step 6: Calculate 2AB.
2AB=2[−3−142]=[−6−284]
Step 7: Calculate A2+2AB+B2.
A2+2AB+B2=[1001]+[−6−284]+[1−304]
=[1−6+10−2−30+8+01+4+4]
=[−4−589]
Step 8: Compare (A+B)(A+B) and A2+2AB+B2.
Since [−2−367]=[−4−589], it is shown that (A+B)(A+B)=A2+2AB+B2.
b) Show that (A+B)(A−B)=A2−B2
Step 1: We already have A+B=[0−123] from part (a).
Step 2: Calculate A−B.
A−B=[−1021]−[1−102]=[−1−10−(−1)2−01−2]=[−212−1]
Step 3: Calculate (A+B)(A−B).
(A+B)(A−B)=[0−123][−212−1]
=[(0)(−2)+(2)(1)(−1)(−2)+(3)(1)(0)(2)+(2)(−1)(−1)(2)+(3)(−1)]
=[25−2−5]
Step 4: We already have A2=[1001] and B2=[1−304] from part (a).
Step 5: Calculate A2−B2.
A2−B2=[1001]−[1−304]
=[1−10−(−3)0−01−4]
=[030−3]
Step 6: Compare (A+B)(A−B) and A2−B2.
Since [25−2−5]=[030−3], it is shown that (A+B)(A−B)=A2−B2.
The final answer is Showninstepsabove.
