Answer:
Step 1
F:43 → \rarr → 43 by
F( \lparen ( z1 ,z2 ,z3 ) \rparen ) = ( \lparen ( 2 z1 +( \lparen ( 1-i) \rparen ) z2 , ( \lparen ( 3+2i) \rparen ) z1 -4iz3 , 2iz1 + ( \lparen ( 4-3i) ) ) z2 -3z3 ) \rparen )
Step 2
Way to find the adjoint of F
( ( ( I) Find matrix of f with respect to basis { \lbrace { e1, e2 , e3 } \rbrace }
where e1 =( ( ( 1,0,0) ) ) , e2 =( ( ( 0,1,0) ) ) , e3 =( ( ( 0,0,1) ) )
Let A be the matrix corresponding to F.
( ( ( II) Compute ( ( ( A T ‾ \overline{ A^T} A T ) ) )
( ( ( II) Hence adjoint F* : 43 → \rarr → 43 is
F* ( \lparen ( z1 ,z2 ,z3 ) \rparen ) = ( ( ( A T ‾ \overline{ A^T} A T ) ) ) [ z 1 z 2 z 3 ] \begin{bmatrix}
z1\\
z2\\
z3\\
\end{bmatrix} ⎣ ⎡ z 1 z 2 z 3 ⎦ ⎤
Now
F( ( ( e1 ) ) ) = F( ( ( 1,0,0) ) ) = ( ( ( 2,( \lparen ( 3+2i) \rparen ) , 2i) ) )
F( ( ( e2 ) ) ) = F( ( ( 0,1,0) ) ) = ( ( ( 1-i, 0, 4-3i) ) )
F( ( ( e3 ) ) ) = F( ( ( 0,0,1) ) ) = ( ( ( 0,-4i, -3) ) )
Step 3
Hence
A= [ 2 1 − i 0 3 + 2 i 0 − 4 i 2 i 4 − 3 i − 3 ] \begin{bmatrix}
2 & 1-i & 0 \\
3+2i & 0 & -4i\\
2i & 4-3i & -3 \\
\end{bmatrix} ⎣ ⎡ 2 3 + 2 i 2 i 1 − i 0 4 − 3 i 0 − 4 i − 3 ⎦ ⎤
AT = [ 2 3 + 2 i 2 i 1 − i 0 4 − 3 i 0 − 4 i − 3 ] \begin{bmatrix}
2 & 3+2i & 2i \\
1-i & 0 & 4-3i \\
0 & -4i & -3 \\
\end{bmatrix} ⎣ ⎡ 2 1 − i 0 3 + 2 i 0 − 4 i 2 i 4 − 3 i − 3 ⎦ ⎤
A T ‾ \overline{ A^T} A T =[ 2 3 − 2 i − 2 i 1 + i 0 4 + 3 i 0 4 i − 3 ] \begin{bmatrix}
2 & 3-2i & -2i \\
1+i & 0 & 4+3i \\
0 & 4i & -3 \\
\end{bmatrix} ⎣ ⎡ 2 1 + i 0 3 − 2 i 0 4 i − 2 i 4 + 3 i − 3 ⎦ ⎤
Hence F* : 43 → \rarr → 43 is
F* ( \lparen ( z1 ,z2 ,z3 ) \rparen ) = ( ( ( A T ‾ \overline{ A^T} A T ) ) ) [ z 1 z 2 z 3 ] \begin{bmatrix}
z1\\
z2\\
z3\\
\end{bmatrix} ⎣ ⎡ z 1 z 2 z 3 ⎦ ⎤
=[ 2 3 − 2 i − 2 i 1 + i 0 4 + 3 i 0 4 i − 3 ] \begin{bmatrix}
2 & 3-2i & -2i \\
1+i & 0 & 4+3i \\
0 & 4i & -3 \\
\end{bmatrix} ⎣ ⎡ 2 1 + i 0 3 − 2 i 0 4 i − 2 i 4 + 3 i − 3 ⎦ ⎤ [ z 1 z 2 z 3 ] \begin{bmatrix}
z1\\
z2\\
z3\\
\end{bmatrix} ⎣ ⎡ z 1 z 2 z 3 ⎦ ⎤
=( \lparen ( 2 z1 +( \lparen ( 3+2i) \rparen ) z2 - 2iz3, ( \lparen ( 1+i) \rparen ) z1 +4+3iz3 , iz2 -3z3 ) \rparen )
Therefore,
F ( \lparen ( z1 ,z2 ,z3 ) \rparen ) =( \lparen ( 2 z1 +( \lparen ( 3+2i) \rparen ) z2 - 2iz3, ( \lparen ( 1+i) \rparen ) z1 +4+3iz3 , iz2 -3z3 ) \rparen )
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