Linear Algebra Solutions 57

Linear Algebra - 9 1 2 i 1 2i 2 R R1 R2 1 i 1 i 1 i 11 R3 R3 R2 i 1 2i 2 i 1 i 1i 1 R2 R2(1 i)R1 0 0 i R2 iR2 R3 R3 iR1 00 0 1i0 R1 R1 R2 0 0 1 000

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9. 2 + i - 1 + 2 i 2 1 + i - 1 + i 1 1 + 2 i - 2 + i 1 + i R 1 R 1 - R 2 R 3 R 3 - R 2 1 i 1 1 + i - 1 + i 1 i - 1 i R 2 R 2 - (1 + i ) R 1 R 3 R 3 - iR 1 1 i 1 0 0 - i 0 0 0 R 2 iR 2 1 i 1 0 0 1 0 0 0 R 1 R 1 - R 2 1 i 0 0 0 1 0 0 0 . The last matrix is in reduced row–echelon form. 10. (i) Let p = l + im and z = x + iy . Then pz + p z = ( l - im )( x + iy ) + ( l + im )( x - iy ) = ( lx + liy - imx + my ) + ( lx - liy + imx + my ) = 2( lx + my ) . Hence pz + p z = 2 n lx + my = n. (ii) Let w be the complex number which results from reFecting the com- plex number z in the line lx + my = n . Then because p is perpendicular to the given line, we have w - z = tp, t R . ( a ) Also the midpoint
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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