Questions

if $A = [\begin{array}{cc} - i & 0 \\ 0 & i \end{array}]$ then A^TA is equal to,

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a

l

b

- iA

c

-l

d

iA

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detailed solution

Correct option is C

$A = [\begin{array}{cc} - i & 0 \\ 0 & i \end{array}], A^{T} = [\begin{array}{cc} - i & 0 \\ 0 & i \end{array}]$
$A^{T} A = [\begin{array}{cc} - i & 0 \\ 0 & i \end{array}] [\begin{array}{cc} - i & 0 \\ 0 & i \end{array}] = [\begin{array}{cc} i^{2} & 0 \\ 0 & i^{2} \end{array}] = [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = - I$

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