Exercise 7.B.15

Answers

Proof. We have

(\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & x \end{matrix}) (\begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & x \end{matrix}) = (\begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & x \end{matrix}) (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & x \end{matrix}) .

Multiplying the third row with the second column on both sides we get

1 \cdot 0 + 0 \cdot 1 + x \cdot 1 = 0 \cdot 1 + 1 \cdot 1 + x \cdot 0,

therefore $x = 1$ . □

celiopassos

2017-10-06 00:00