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Question Number 81879 by zainal tanjung last updated on 16/Feb/20
The value of determinant determinant (((−1),(    1),(    1)),((    1),(−1),(    1)),((    1),(     1),(−1)))  is equal to
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{determinant}\begin{vmatrix}{−\mathrm{1}}&{\:\:\:\:\mathrm{1}}&{\:\:\:\:\mathrm{1}}\\{\:\:\:\:\mathrm{1}}&{−\mathrm{1}}&{\:\:\:\:\mathrm{1}}\\{\:\:\:\:\mathrm{1}}&{\:\:\:\:\:\mathrm{1}}&{−\mathrm{1}}\end{vmatrix} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by john santu last updated on 16/Feb/20
−1(0)−1(−2)+1(2) = 4
$$−\mathrm{1}\left(\mathrm{0}\right)−\mathrm{1}\left(−\mathrm{2}\right)+\mathrm{1}\left(\mathrm{2}\right)\:=\:\mathrm{4} \\ $$
Commented by mathmax by abdo last updated on 16/Feb/20
=−1× determinant (((−1        1)),((1            −1)))−1× determinant (((1         1)),((1         −1)))+1× determinant (((1       1)),((−1   1)))  =−(0)−(−2)+(2)  =4
$$=−\mathrm{1}×\begin{vmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{vmatrix}−\mathrm{1}×\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{vmatrix}+\mathrm{1}×\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{1}\:\:\:\mathrm{1}}\end{vmatrix} \\ $$$$=−\left(\mathrm{0}\right)−\left(−\mathrm{2}\right)+\left(\mathrm{2}\right)\:\:=\mathrm{4} \\ $$

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