STA3710 Assignment 4 (894289) |COMPLETE WORKINGS & SOLUTIONS|- DUE 9 September 2025

STA3710
Assignment 4

Unique No: 894289
DUE 9 September 2025

, ASSIGNMENT 04
Unique Nr.: 894289
Fixed due date: 9 September 2025


Question 1.1

𝑨𝟏𝟏 𝑨𝟏𝟐
Given a matrix 𝑨 partitioned as 𝑨 = ( ), where 𝑨𝟏𝟏 is an 𝒎𝟏 × 𝒎𝟏 matrix.
𝑨𝟐𝟏 𝑨𝟐𝟐
We're also given that 𝒓𝒂𝒏𝒌(𝑨) = 𝒓𝒂𝒏𝒌(𝑨 𝟏𝟏 ) = 𝒎 𝟏 .

We need to show that 𝑨𝟐𝟐 = 𝑨 𝟐𝟏 𝑨−𝟏
𝟏𝟏 𝑨𝟏𝟐 .


A key property of partitioned matrices is that if 𝑟𝑎𝑛𝑘(𝐴) = 𝑟𝑎𝑛𝑘(𝐴 11 ), it implies that the
row and column spaces of 𝐴 are fully determined by the submatrix 𝐴11 and its
corresponding rows and columns.

Specifically, the rows of the block 𝐴21 are linear combinations of the rows of 𝐴11 , and
similarly for the columns. This means that the block 𝐴22 can be expressed in terms of
the other blocks.

The condition 𝑟𝑎𝑛𝑘(𝐴) = 𝑟𝑎𝑛𝑘(𝐴 11 ) implies that the Schur complement of 𝐴11 , which is
−1
𝑆 = 𝐴 22 − 𝐴21 𝐴11 𝐴12 , must be a zero matrix.

Given the condition 𝑟𝑎𝑛𝑘(𝐴) = 𝑟𝑎𝑛𝑘(𝐴 11 ) = 𝑚 1 , the Schur complement of 𝐴11 is zero.
−1
The Schur complement of 𝐴11 is defined as 𝐴22 − 𝐴21 𝐴11 𝐴12 .

−1
Setting the Schur complement to zero, we get: 𝐴22 − 𝐴21 𝐴11 𝐴12 = 0

−1
Rearranging the terms, we get: 𝐴22 = 𝐴 21 𝐴11 𝐴12 .

Question 1.2

𝒂𝑰𝒎 𝒃𝑰𝒎
Consider the matrix 𝑨 = ( ), where 𝒂, 𝒃, 𝒄, 𝒅 are scalars. We're given that
𝒄𝑰𝒎 𝒅𝑰𝒎
𝒎𝟏 = 𝒎 𝟐 and 𝑨𝟏𝟏 𝑨𝟐𝟏 = 𝑨 𝟐𝟏 𝑨𝟏𝟏 .

We need to prove that |𝑨| = |𝑨 𝟏𝟏 𝑨𝟐𝟐 − 𝑨𝟐𝟏 𝑨𝟏𝟐 |.


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