HP 15c Scientific Solving the Complex Equation AX = B

168 Section 12: Calculating with Matrices

Writing down the elements of C,

⎡	1.0000	− 2.8500×10−10		⎤
⎢	−11			⎥	= (ZZ −1 )P
C = ⎢− 4.0000×10−		1.0000	−	⎥		,
⎢	1.0000×10 11	3.8000×10 10		⎥
⎢	1.0000×10−11	− 1.0500×10−10		⎥
⎣⎢				⎦⎥

where the upper half of matrix C is the real part of ZZ-1and the lower half is the imaginary part. Therefore, by inspection of matrix C,

ZZ −1	⎡	1.0000	− 2.8500×10	−10 ⎤
	= ⎢			⎥
	⎣⎢− 4.0000×10−11		1.0000	⎦⎥

⎡1.0000×10−11		3.8000×10−11	⎤
+ i ⎢	−11	−10	⎥
⎣⎢1.0000×10		−1.0500×10	⎦⎥

As expected,

ZZ -1	⎡1	0⎤	⎡0	0⎤
	= ⎢	⎥	+ i ⎢	⎥
	⎣0	1⎦	⎣0	0⎦

Solving the Complex Equation AX = B

You can solve the complex matrix equation AX = B by finding X = A-1B. Do this by calculating XP = (Ã)-1BP.

To solve the equation AX = B, where A, X, and B are complex matrices:

1.Store the elements of A and B in memory, in the form either of ZP or of ZC.

2.Recall the descriptor of the matrix representing B into the display.

3.If the elements of B were entered in the form BC, press ´p to transform BC into BP.