Chapter 2.5, Problem 8P

Explanation of Solution

a.

Inverse of the given matrix:

We know that for any square matrix $A$,

$A^{-1}A=I$

Therefore, we have,

${(100B)}^{-1}(100B)=I$

Hence, we get

Explanation of Solution

b.

Obtaining the inverse:

Suppose $B\text{'}$ be the matrix obtained by doubling every entry in row 1 of any $n\times n$ matrix $B$.

Then we have,

$B\text{'}=\left[\begin{array}{l}\text{2 0 }\cdots \text{ 0}\\ \text{0 1 }\cdots \text{ 0}\\ ⋮⋮\ddots ⋮\\ \text{0 0 }\cdots \text{ 1}\end{array}\right]B$

Now, we have,

${(B\text{'})}^{-1}={\left(\left[\begin{array}{l}\text{2 0 }\cdots \text{ 0}\\ \text{0 1 }\cdots \text{ 0}\\ ⋮⋮\ddots ⋮\\ \text{0 0 }\cdots \text{ 1}\end{array}\right]B\right)}^{-1}$

  

Explanation of Solution

c.

Obtaining the inverse:

Suppose $B\text{'}$ be the matrix obtained by doubling every entry in column 1 of any $n\times n$ matrix $B$.

Then, we have,

$B\text{'}=B\left[\begin{array}{l}\text{2 0 }\cdots \text{ 0}\\ \text{0 1 }\cdots \text{ 0}\\ ⋮⋮\ddots ⋮\\ \text{0 0 }\cdots \text{ 1}\end{array}\right]$

Now, we have,

${(B\text{'})}^{-1}={\left(B\left[\begin{array}{l}\text{2 0 }\cdots \text{ 0}\\ \text{0 1 }\cdots \text{ 0}\\ ⋮⋮\ddots ⋮\\ \text{0 0 }\cdots \text{ 1}\end{array}\right]\right)}^{-1}$