Summary MAT1503 ASSIGNMENT 2 2024

MAT1503
ASSIGNMENT 2
2024

, QUESTION 1




Solution:



1.1).



At (Dt B)t − αB(CDt )t



Dt has the size (5 × 3) and B has the size (4 × 5) , therefore Dt B is not defined. The
matrix expression is not defined because Dt B is not defined.



1.2).



(B(X t − I)t A)t + B = At

, (B(X t − I)t A)t = At − B
((B(X t − I)t A)t )t = (At − B)t

B(X t − I)t A = (At − B)t
B −1 B(X t − I)t A = B −1 (At − B)t
(X t − I)t A = B −1 (At − B)t

(X t − I)t AA−1 = B −1 (At − B)t A−1
(X t − I)t = B −1 (At − B)t A−1

((X t − I)t )t = (B −1 (At − B)t A−1 )t

X t − I = (B −1 (At − B)t A−1 )t

X t = (B −1 (At − B)t A−1 )t + I
(X t )t = ((B −1 (At − B)t A−1 )t + I)t

X = ((B −1 (At − B)t A−1 )t + I)t




1.3).



A(A−1 + B −1 )B = A + B
(AA−1 + AB −1 )B = A + B

(I + AB −1 )B = A + B

IB + AB −1 B = A + B

B + AI = A + B
B+A =A+B
A+B =A+B (proven)



1.4).



Let A be a square matrix such that

A3 + 4A2 − 2A + 7In = 0



The by properties of transpose:
0 = 0t