Question

In this exercise we show that the meet and join operations are commutative. Let A and B be $m\times n$zero-one matrices. Show that

a) $A\vee B=B\vee A$

b)$B\wedge A=A\wedge B$

Short answer

Hence, proved

1. $A\vee B=B\vee A$.
2. $B\wedge A=A\wedge B$
   

Step-by-step solution

Step 1:

Definition of matrix:

A matrix is a rectangular array of numbers. A matrix withrows andcolumns is called as$m\times n$ matrix. :

Step 2:

(a)

A and B are $m\times n$zero-one matrix

$A\vee B=B\vee A$

Proof: We can rewrite the matrix A as$\left[a_{ij}\right]$ where$a_{ij}$ represents the element in the ith row and in the jth column of A.

$\begin{array}{rr}A\vee B& \end{array}$Takes disjunction of each element of A with the corresponding element

$\begin{array}{r}A\vee B=\left[a_{ij}\right]\vee \left[b_{ij}\right]\\ =\left[a_{ij}\vee b_{ij}\right]\end{array}$

Use commutative law

$$\begin{gathered} =[b_{i}j\vee a_{i}j] \\ =B\vee A \end{gathered}$$

Step 3:

(b)

A and B are $m\times n$zero-one matrix

$B\wedge A=A\wedge B$

Proof: We can rewrite the matrix A as $\left[a_{ij}\right]$where $a_{ij}$represents the element in the $i^{th}$ row and in the$j^{th}$ column of A.

$\mathrm{B}\wedge \mathrm{A}$,takes conjunction of each element of A with the corresponding element

$\begin{array}{r}\mathrm{B}\wedge \mathrm{A}=\left[{\mathrm{b}}_{\mathrm{ij}}\right]\wedge \left[{\mathrm{a}}_{\mathrm{ij}}\right]\\ =\left[{\mathrm{b}}_{\mathrm{ij}}\wedge {\mathrm{a}}_{\mathrm{ij}}\right]\end{array}$

Use commutative law

localid="1668492289626" $$\begin{gathered} =[a_{i}j\wedge b_{i}j] \\ =A\wedge B \end{gathered}$$

Hence, the commutative laws are proved.