Chapter 2.1, Problem 1P

Explanation of Solution

a.

Finding -A:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]andB=[\begin{array}{l}\text{1 2}\\ \end{array}$

Explanation of Solution

b.

Finding 3A:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]andB=\left[\begin{array}{l}\text{1 2}\\ \text{0 -1}\\ \text{1 2}\end{array}\right]$

In order to find matrix 3A, multiply the given matrix A by scalar 3.

Therefore, the matrix obtained is as follows:

$\begin{array}{l}\text{3A=3}\times [\begin{array}{l}\text{1 2 x}\end{array}\end{array}$

Explanation of Solution

c.

Finding A+2B:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]andB=\left[\begin{array}{l}\text{1 2}\\ \text{0 -1}\\ \text{1 2}\end{array}\right]$

In order to find matrix 2B, multiply the given matrix B by scalar 2

Explanation of Solution

d.

Finding Transpose of matrix, A:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]andB=[\begin{array}{l}\text{1 2}\\ \text{0 -1}\\ \text{1 x}\end{array}$

Explanation of Solution

e.

Finding Transpose of matrix, B:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]andB=[\begin{array}{l}\text{1 2}\\ \text{0 -1}\end{array}$

Explanation of Solution

f.

Finding AB:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]andB=\left[\begin{array}{l}\text{1 2}\\ \text{0 -1}\\ \text{1 2}\end{array}\right]$

Here, the number of columns in A are equal to the number of rows in B, therefore, the multiplication AB exists. Thus, the matrix AB will be of the order 3x2.

Now, suppose C=AB.

Therefore,

$\begin{array}{l}\text{C=AB=}\left[\begin{array}{l}\text{1 4 7}\\ \text{2 5 8}\\ \text{3 6 9}\end{array}\right]\left[\begin{array}{l}\text{1 2}\\ \text{0 -1}\\ \text{1 2}\end{array}\right]\\ \end{array}$

$\begin{array}{l}{\text{C}}_{\text{11}}\text{=}\left[\text{1 2 3}\right]\left[\begin{array}{l}\text{1}\\ \text{0}\\ \text{1}\end{array}\right]\\ \text{ =1(1)+2(0)+3(1)}\\ \text{ =1+0+3}\\ \text{ =4}\\ \end{array}$

$\begin{array}{l}{\text{C}}_{\text{12}}\text{=}\left[\text{1 2 3}\right]\left[\begin{array}{l}\text{ 2}\\ \text{-1}\\ \text{ 2}\end{array}\right]\\ \text{ =1(2)+2(-1)+3(2)}\\ \text{ =2+(-2)+6}\\ \text{ =6}\end{array}$

$\begin{array}{l}{\text{C}}_{\text{2}}\end{array}$

Explanation of Solution

g.

Finding BA:

Consider the given matrices,

$\text{A=}\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ \text{7 8 9}\end{array}\right]a$