Question

show that matrix
| 1 a b ab |
| 1 a' b a'b |
| 1 a b' ab' |
| 1 a' b' a'b |
+
= (a'-a)(b'-b)^2

Answer 1

are you sure the (4,4) element is correct shouldn't it be a'b'

Answer 2

yes

Answer 3

what is the problem ?

Answer 4

$$\left|\begin{matrix}
1 & a & b & ab\\
1 & a' & b & a'b\\
1 & a & b' & ab'\\
1 & a' & b' & a'b
\end{matrix}\right|\xrightarrow[R_2-R_4\rightarrow R_2]{R_1-R_3\rightarrow R_1}$$
$$\left|\begin{matrix}
0 & 0 & (b-b') & a(b-b')\\
0 & 0 & (b-b') & 0\\
1 & a & b' & ab'\\
1 & a' & b' & a'b
\end{matrix}\right|=(b-b')^2\left|\begin{matrix}
0 & 0 & 1 & a\\
0 & 0 & 1 & 0\\
1 & a & b' & ab'\\
1 & a' & b' & a'b
\end{matrix}\right|=-(b-b')^2\left|\begin{matrix}
0 & 0 & 1 & 0\\
0 & 0 & 1 & a\\
1 & a & b' & ab'\\
1 & a' & b' & a'b
\end{matrix}\right|=$$
$$-(b-b')^2\left|\begin{matrix}
0 & 0 & a \\
1 & a & ab'\\
1 & a' & a'b
\end{matrix}\right|=-(b-b')^2\left[a(a'-a)\right]=a(a-a')(b-b')^2 $$This is what I get. Please check If there is a mistake

Answer 5

thanks for your help

Answer 6

you're welcome :)