Summary
Summary MAT1503 ASSIGNMENT 2 2024
- Course
- MAT1503 (MAT1503)
- Institution
- University Of South Africa (Unisa)
This document contains MAT1503 ASSIGNMENT 2 2024 solutions. Clear step by step calculation are provided.
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MAT1503ASSIGNMENT 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
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