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MAT1503 Linear Algebra Assignment 2
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- University Of South Africa (Unisa)
Full solutions to assignment 2 with proofs
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Name: Phillip Martin Le GrangeContact No. 0720494056
Student Number: 19210590
Subject: MAT1503
Assessment: 2
Question 1
A - (4 × 1), B - (4 × 5), C - (3 × 3), D (3 × 5)
AT – (1x4), BT – (5 x 4), CT – (3x3), DT – (5 x 3)
(i) BT(A+DT)
Cannot add A and DT and while BTA is defined BTDT is not. Therefore, this expression does
not result in a defined matrix
(ii) A(DBT)+2DTB
DTB is not defined as they cannot be multiplied together. DBT is defined as they can be
multiplied but A(DBT) is also not defined. Therefore, this expression does not result in a
defined matrix.
(iii) (CD)BT+λAC
AC is not defined (cannot multiply) but (CD)BT is. Therefore, this expression does not
result in a defined matrix.
,Question 2
2.1. (𝐵(𝑋 𝑡 − 𝐼)−1 𝐴)𝑡 + 𝐵 = 𝐴𝑡
(𝐵(𝑋 𝑡 − 𝐼)−1 𝐴)𝑡 + 𝐵 − 𝐵 = 𝐴𝑡 − 𝐵
((𝐵(𝑋 𝑡 − 𝐼)−1 𝐴)𝑡 )𝑡 = (𝐴𝑡 − 𝐵)𝑡
𝐵 −1 (𝐵(𝑋 𝑡 − 𝐼)−1 𝐴)𝐴−1 = 𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1
((𝑋 𝑡 − 𝐼)(𝑋 𝑡 − 𝐼)−1 ) = (𝑋 𝑡 − 𝐼)(𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )
𝐼 = (𝑋 𝑡 − 𝐼)(𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )
𝐼 + (𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 ) = (𝑋 𝑡 )(𝐵−1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )
𝐼(𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )−1 + (𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )(𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )−1
= (𝑋 𝑡 )(𝐵−1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )(𝐵−1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )−1
(𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 )−1 + 𝐼 = 𝑋 𝑡
[(𝑩−𝟏 (𝑨𝒕 − 𝑩)𝒕 𝑨−𝟏 )−𝟏 + 𝑰]𝒕 = 𝑿
2.2. (𝐵(𝑋 −1 − 𝐼)𝐴)𝑡 + 𝐵 = 𝐴𝑡
(𝐵(𝑋 −1 − 𝐼)𝐴)𝑡 = 𝐴𝑡 − 𝐵
(𝐵(𝑋 −1 − 𝐼)𝐴) = (𝐴𝑡 − 𝐵)𝑡
(𝑋 −1 − 𝐼) = 𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1
(𝑋 −1 ) = 𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 + 𝐼
(𝑋)(𝑋 −1 ) = (𝑋)[𝐵−1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 + 𝐼]
𝐼 = (𝑋)[𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 + 𝐼]
𝐼 = (𝑋)[𝐵 −1 (𝐴𝑡 − 𝐵)𝑡 𝐴−1 + 𝐼]
[𝑩−𝟏 (𝑨𝒕 − 𝑩)𝒕 𝑨−𝟏 + 𝑰]−𝟏 = 𝑿
, 2.3. 𝑋 2 = −5𝑋 + 2𝐼
𝑋𝑋 = −5𝑋 + 2𝐼
𝑋 −1 𝑋𝑋 = −5𝑋 −1 𝑋 + 2𝑋 −1
𝑋 = −5𝐼 + 2𝑋 −1
𝟏 𝟓
𝑿 + 𝑰 = 𝑿−𝟏
𝟐 𝟐
While X is an invertable matrix nothing is said of Y. The product of invertable matrixes does
result in an invertable matrix.
Thus in the event that Y is not invertable (In+YX-1) is also not invertable.
The sum of X + Y is inconclusive and may be invertable or not depending on the value of Y.
𝑋(𝑋 −1 + 𝑌 −1 )𝑌(𝑋 + 𝑌)−1 = 𝐼
Multiply X into first bracket
(𝐼 + 𝑋𝑌 −1 )𝑌(𝑋 + 𝑌)−1 = 𝐼
Multiply first bracket with Y
(𝑌 + 𝑋)(𝑋 + 𝑌)−1 = 𝐼
Reorder first bracket and multiply remainder to obtain Identity matrix
(𝑋 + 𝑌)(𝑋 + 𝑌)−1 = 𝐼
𝐼=𝐼
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