Question

Let be a zero–one matrix. Show that

a) $\mathit{A}\mathbf{\vee }\mathit{A}\mathbf{=}\mathit{A}$

b)$\mathit{A}\mathbf{\wedge }\mathit{A}\mathbf{=}\mathit{A}$

Short answer

Therefore, it is proved that$\mathit{A}\mathbf{\vee }\mathit{A}\mathbf{=}\mathit{A}$ and$\mathit{A}\mathbf{}\mathit{\Lambda }\mathbf{}\mathit{A}\mathbf{=}\mathit{A}$

Step-by-step solution

Step 1:

(a) Given: A is a zero-one matrix.

To prove:$\mathit{A}\mathbf{\vee }\mathit{A}\mathbf{=}\mathit{A}$

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

$\mathit{A}\mathbf{\vee }\mathit{B}$takes the disjunction of each element of A with the corresponding element of B

$A\vee A=\left[a_{i}j\right]\vee \left[a_{i}j\right]=[a_{i}j\vee a_{i}j]$

Use idempotent law:

$\mathbf{=}\mathbf{\left[}{\mathit{a}}_{\mathbf{i}\mathbf{j}}\mathbf{\right]}\mathbf{=}\mathit{A}$

Step 2:

(b) Given: A is a zero-one matrix.

To prove:$\mathit{A}\mathbf{\wedge }\mathit{A}\mathbf{=}\mathit{A}$

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.

$\mathit{A}\mathit{\Lambda }\mathit{B}$takes the conjunction of each element of A with the corresponding element of B.

$A\wedge A=\left[a_i\right]\wedge \left[a_{ij}\right]=[a_{ij}\wedge a_{ij}]$

Use idempotent law:

$=\left[a_{ij}\right]=A$