Doubting Logic: Boolean Matrix Homework Help

[noreturn2]

Homework Statement

Homework Equations

The Attempt at a Solution

Does my logic seem right, I'm doubtin my anwsers.

 

[Mark44]

Homework Statement

View attachment 235016

Homework Equations

The Attempt at a Solution

View attachment 235017
Does my logic seem right, I'm doubtin my anwsers.

Your answers for parts a and b seem OK, but I didn't check that closely, plus you didn't show how the bitwise AND and OR operations are supposed to work.
Your answer to c is incorrect, at least based on how matrix multiplication is normally defined. For the product CC, the upper left entry in the product matrix should be 2, not 1 as you show. This is obtained by calculating the dot product of row 1 of the left matrix with column 1 of the right matrix.

 

[noreturn2]

Boolean values can only be 1 or 0.

I think you are right that it is still wrong.

I re did it and I got:
[ 1 1 1
0 1 0
1 1 1]

 

[Mark44]

Boolean values can only be 1 or 0.

I think you are right that it is still wrong.

I re did it and I got:
[ 1 1 1
0 1 0
1 1 1]

I get something else. Here's what I get for $C^2$.
$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$
$= \begin{bmatrix}0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}$
The entry in the upper left corner deserves some explanation. The 0 value there is the result of the dot product of row 1 on the left (<1, 0, 1>) with col. 1 on the right (<1, 0, 1>^T). From this I get 1*1 + 0*0 + 1*1 = 2, or 10₂. Since this is 2 bits, it won't fit into 1 bit, we get 0 for the upper left corner of the product.
I'm assuming that's how you're supposed to be doing multiplication of Boolean values.

 

[noreturn2]

I get something else. Here's what I get for $C^2$.
$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$
$= \begin{bmatrix}0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}$
The entry in the upper left corner deserves some explanation. The 0 value there is the result of the dot product of row 1 on the left (<1, 0, 1>) with col. 1 on the right (<1, 0, 1>^T). From this I get 1*1 + 0*0 + 1*1 = 2, or 10₂. Since this is 2 bits, it won't fit into 1 bit, we get 0 for the upper left corner of the product.
I'm assuming that's how you're supposed to be doing multiplication of Boolean values.

Well if you have 1*1 on any column that value is autoamtically 1. So if you like at row 1 column one on each that is why I get a 1 in the corner

 

[Mark44]

Well if you have 1*1 on any column that value is autoamtically 1. So if you like at row 1 column one on each that is why I get a 1 in the corner

I agree that 1*1 = 1, but I was talking about 1*1 + 0*0 + 1*1 = 1 + 1 = 2.
That's how ordinary matrix multplication would be done, but you didn't include any information about how the Boolean values are to be added. For example, 0 + 0 = 0, 1 + 0 = 0 + 1 = 1, but what is 1 + 1? In terms of bits, which are essentially Boolean values, 1 + 1 = 0, with a carry of 1.

 

[haruspex]

I do not understand how you calculated part b.
E.g. how do you get that 0 in the second column, first row?