First slide

Transpose of matrix

Question

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

Easy

A

I
B

-iA
C

-I
D

iA

Solution

We have
$A^{'} A = [\begin{array}{rr} - i 0 \\ 0 i \end{array}] [\begin{array}{rr} - i 0 \\ 0 i \end{array}] = [\begin{array}{rr} - 1 0 \\ 0 - 1 \end{array}] = - I$

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