MATH 122 – H Activity 2
Linear Algebra and Matrix Theory Regine Joyce A. Camacho – BS Mathematics
I. MODIFIED TRUE OR FALSE.
1. FALSE. Type 3
2. FALSE. Type 2
3. FALSE. Only trivial solution
4. TRUE
5. FALSE. Has no zero row.
II. COMPUTATIONS.
1. Let 𝐴=[2 −1
3 2] and 𝐵=[−4 2
1 3]. Using these matrices, show that
a. (𝐴+𝐵)2≠𝐴2+2𝐴𝐵+𝐵2
Solution:
Given that 𝐴=[2 −1
3 2] and 𝐵=[−4 2
1 3], 𝐴+𝐵 is
[2 −1
3 2]+[−4 2
1 3]=[−2 1
4 5]
With that, (𝐴+𝐵)2=(𝐴+𝐵)(𝐴+𝐵) is
[−2 1
4 5][−2 1
4 5]=[(−2)(−2)+(1)(4) (−2)(1)+(1)(5)
(4)(−2)+(5)(4) (4)(1)+(5)(5)]
=[4+4 −2+5
−8+20 4+25]
⇒(𝐴+𝐵)2=[8 3
12 29]
Since 𝐴=[2 −1
3 2], this implies that 𝐴2=(𝐴)(𝐴) is
[2 −1
3 2][2 −1
3 2]=[(2)(2)+(−1)(3) (2)(−1)+(−1)(2)
(3)(2)+(2)(3) (3)(−1)+(2)(2)]
=[4−3 −2−2
6+6 −3+4]
⇒𝐴2=[1 −4
12 1]
Since 𝐴=[2 −1
3 2] and 𝐵=[−4 2
1 3], this implies that 𝐴𝐵 is
[2 −1
3 2][−4 2
1 3]=[(2)(−4)+(−1)(1) (2)(2)+(−1)(3)
(3)(−4)+(2)(1) (3)(2)+(2)(3)]
=[−8−1 4−3
−12+2 6+6]
⇒𝐴𝐵=[−9 1
−10 12]
With that, 2𝐴𝐵 is 2[−9 1
−10 12]=[−18 2
−20 24]
