Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a $1\times 1$ matrix,

$A=\left[a\right]$

Then, $\det (A)=a$

Now, we have

$-A=\left[-a\right]$

Then, we get

$\begin{array}{l}\det (-A)=-a\\ =-\det (A)\end{array}$

Similarly consider a $3\times 3$ matrix as given below:

$A=[\begin{array}{l}a\text{ b c}\\ 0\text{ d e}\\ 0\text{ f }\end{array}$

Explanation of Solution

b.

Proof:

Let us consider a  $2\times 2$ matrix as given below:

$A=\left[\begin{array}{l}a\text{ b}\\ c\text{ d}\end{array}\right]$

Then, we get $\det (A)=ad-bc$

Now, we have

$-A=\left[\begin{array}{l}-a\text{ -b}\\ -c\text{ -d}\end{array}\right]$

Then, we get

$\begin{array}{l}\det (-A)=ad-bc\\ =\text{det(A)}\end{array}$

Similarly consider a $4\times 4$ matrix as,

$A=[\begin{array}{l}a\text{ b c d}\\ 0\text{ e f g}\\ 0\text{ h 0 �}\end{array}$

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any $n\times n$ matrix,

$\begin{array}{l}\det (-A)=-\det (\end{array}$