- #1
jwqwerty
- 43
- 0
if A B C D are nxn matrices such that rank(A)=rank([A B])=n ([A B] : 2nx2n matrix)
[C D] [C D]
we want to show that D=CA‑¹B (A‑¹: inverse matrix of A)
this is what i have tried:
[In 0] [A B] = [A B ] (here, In refers to identity nxn matrix)
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]
det( [In 0] [A B] ) = det ( [A B ] )
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]
det( [In 0] ) det( [A B] ) = det ([A B ] )
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]
by property of determinant of block matrix,
det(In)det(In)det( [A B] ) = det(A)det(CA‑¹B+D)
[C D]
since A is invertible and [A B] is not invertible, their determinents are not zero and zero respectively
[C D]
Thus, det(CA‑¹B+D)=0
If i can show that CA‑¹B+D=0, then my proof ends.
But i can't show this.
can anyone show that CA‑¹B+D=0?
[C D] [C D]
we want to show that D=CA‑¹B (A‑¹: inverse matrix of A)
this is what i have tried:
[In 0] [A B] = [A B ] (here, In refers to identity nxn matrix)
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]
det( [In 0] [A B] ) = det ( [A B ] )
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]
det( [In 0] ) det( [A B] ) = det ([A B ] )
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]
by property of determinant of block matrix,
det(In)det(In)det( [A B] ) = det(A)det(CA‑¹B+D)
[C D]
since A is invertible and [A B] is not invertible, their determinents are not zero and zero respectively
[C D]
Thus, det(CA‑¹B+D)=0
If i can show that CA‑¹B+D=0, then my proof ends.
But i can't show this.
can anyone show that CA‑¹B+D=0?