Solution: det (A) = −5, and for n×n matrix adj (A) has determinant (det A)^ (n−1). Here n = 3, so det (B) = (−5)^ (2) = 25.
In the Hilbert space L A 2 ( I;E )( I:=[ a,b ),−∞ 0 ) , the maximal dissipative singular matrix-valued Sturm–Liouville operators that the extensions of a minimal symmetric operator with maximal ...
Results that may be inaccessible to you are currently showing.
Hide inaccessible results
Feedback