Chapter 8.4, Problem 29E

a.

To determine

To calculate: Minors

Answer to Problem 29E

  $M_{11}=-6,M_{12}=3,M_{21}=5,M_{22}=4$

Explanation of Solution

Given information: Matrix is $\left[\begin{array}{cc}4& 5\\ 3& -6\end{array}\right]$

Calculation:

Let $M=\left[\begin{array}{cc}4& 5\\ 3& -6\end{array}\right]$

First find minors:

For $M_{11}$ , remove first row and first column. S, the remaining element is -6.

So, $M_{11}=-6$

For $M_{12}$ , remove first row and second column. S, the remaining element is 3.

So, $M_{12}=3$

For $M_{21}$ , remove second row and first column. S, the remaining element is 5.

So, $M_{21}=5$

For $M_{22}$ , remove second row and second column. S, the remaining element is 4.

So, $M_{22}=4$

Therefore,

  $M_{11}=-6,M_{12}=3,M_{21}=5,M_{22}=4$

b.

To determine

To calculate: cofactors

Answer to Problem 29E

  $\begin{array}{l}{A}_{11}=-6\\ A_{12}=-3\\ A_{21}=-5\\ A_{22}=4\end{array}$

Explanation of Solution

Formula used: If $M_{ij}$ then cofactors are given by $A_{ij}={(-1)}^{i+j}{M}_{ij}$

Calculation:

As calculated in previous part, minors are $M_{11}=-6,M_{12}=3,M_{21}=5,M_{22}=4$

Now find cofactors using the formula $A_{ij}={(-1)}^{i+j}{M}_{ij}$

  $A_{11}={(-1)}^{1+1}{M}_{11}=1(-6)=-6$

  $A_{12}={(-1)}^{1+2}{M}_{12}=-1(3)=-3$

  $A_{21}={(-1)}^{2+1}{M}_{21}=-1(5)=-5$

  $A_{22}={(-1)}^{2+2}{M}_{22}=1(4)=4$