Question

Let A be ann x nzero-one matrices. Let I be theidentity matrix. Show that $A\odot I=I\odot A=A$

Short answer

Hence, proved

$A\odot I=I\odot A=A$.

Step-by-step solution

Step 1:

definition of matrix:

A matrix is a rectangular array of numbers. A matrix with m rows and n columns is called as m x n matrix. :

Identity law for propositions:

$$\begin{gathered} p\wedge T\equiv p \\ p\vee F\equiv p \end{gathered}$$

Step 2

Given: A is a n x n zero-one matrix and I is theidentity matrix

To proof:$A\odot I=I\odot A=A$

First part we can rewrite the matrix A aswhererepresents the element in the ith row and in the jth column of A. we name as the elements of I.

$A\odot I=\left[a_{i}j\right]\odot \left[b_{i}j\right]$

The ijth element of the Boolean product Takes the disjunction of all conjunctions of corresponding elements A and B in the ith row of A and in the jth column of B.

$=\left[\underset{q}{\underbrace{V}}(a_{i}q\wedge b_{q}j)\right]$

Sinceare the elements of the identity matrix: andwhen. represents true T andrepresents false F.

$=\left[\underset{T}{\underbrace{V}}┬(q\ne j)(a_{i}q\wedge F)\vee (a_{i}j\wedge T)\right]$

Use identity law

$=\left[\underset{T}{\underbrace{V}}(q\ne j)(a_{i}q\wedge F)\vee a_{i}j\right]$

Use domination law

$=\left[\underset{T}{\underbrace{V}}┬(q\ne j)F\vee a_{i}j\right]$

Use idempotent law

$=[F\vee a_{i}j]$

Use identity law

$$\begin{gathered} =\left[a_{i}j\right] \\ =A \end{gathered}$$

Step 3

Second part we can rewrite the matrix A aswhererepresents the element in the ith row and in the jth column of A. we nameas the elements of I.

$I\odot A=\left[b_{i}j\right]\odot \left[a_{i}j\right]$

The ijth element of the Boolean product$A\odot B$ Takes the disjunction of all conjunctions of corresponding elements A and B in the ith row of A and in the jth column of B.

$=\left[\underset{q}{\underbrace{V}}\left(b_{iq}\wedge a_{qj}\right)\right]$

Sinceare the elements of the identity matrix:$b_{ii}=1$ and$b_{iq}=0$ when.represents true T andrepresents false F.

$=\left[\underset{q\ne i}{\underbrace{V}}\left(F\wedge a_{iq}\right)\vee \left(T\wedge a_{ii}\right)\right]$

Use identity law

$=\left[\underset{q\ne j}{\underbrace{V}}\left(F\wedge a_{qj}\right)\vee a_{ij})\right]$

Use domination law

$=\left[\underset{q\ne j}{\underbrace{V}}F\vee a_{ij}\right]$

Use idempotent law

$=[F\vee a_{i}j]$

Use identity law

$$\begin{gathered} =\left[a_{i}j\right] \\ =A \end{gathered}$$

Hence, the identity matrix is proved.