Chunk #376 — Appendix

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Review on solving the inverse problem in EEG source analysis.

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K∗KK∗+αK∗IN=K∗KK∗+αIpK∗⇒(K∗KK∗+αIN)=(K∗K+αIp)K∗⇒(K∗K+αIp)−1K∗(KK∗+αIN)(KK∗+αIN)−1=(K∗K+αIp)−1(K∗K+αIp)K∗(KK∗+αIN)−1⇒(K∗K+αIp)−1K∗IN=IpK∗(KK∗+αIN)−1⇒(K∗K+αIp)−1K∗=K∗(KK∗+αIN)−1.