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Question

If $A equals open curly brackets table attributes columnalign left end attributes row i 0 row 0 cell negative i end cell end table close curly brackets$ then A² $equals$

$\{\begin{array}{l} 1 & 0 \\ 0 & - 1 \end{array}\}$
$\{\begin{array}{l} - 1 & 0 \\ 0 & - 1 \end{array}\}$
$\{\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}\}$
$\{\begin{array}{l} - 1 & 0 \\ 0 & 1 \end{array}\}$

hint Hint:

Recall the method of multiplying 2x2 matrices and the value of imaginary number.

The correct answer is: $open curly brackets table attributes columnalign left end attributes row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close curly brackets$

$A equals open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets A squared equals open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets A squared equals open square brackets table row cell i squared plus 0 end cell cell 0 plus 0 end cell row cell 0 plus 0 end cell cell 0 plus open parentheses negative i close parentheses squared end cell end table close square brackets A squared equals open square brackets table row cell i squared end cell 0 row 0 cell i squared end cell end table close square brackets A squared equals open square brackets table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close square brackets space space space space space space space space space space space space open parentheses because i squared equals negative 1 close parentheses$

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