# Show that if the matrix equation XS = B can be solved for X , then X = BS −1 is a solution for any…

1. Show that if the matrix equation*XS *= *B *can be solved for*X*, then*X *= *BS*^{−1} is a solution for *any *generalized inverse *S*−1 of *S*.

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Show that if the matrix equation XS = B can be solved for X , then X = BS −1 is a solution for any…

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2. Use the result of (a) to derive the expression for *P(*X|Y*)*

3. Verify that the steady-state solution of the Kalman recursions is given by