Answer:
d) Not necessarily
e) True
f)
i) AB is non singular
ii) AB is singular
iii) AB is singular
g) Not necessarily
Step-by-step explanation:
d) Not neccesarily, if we take
![A = \left[\begin{array}{cc}1&1\\1&2\end{array}\right] \\B = \left[\begin{array}{cc}0&2\\2&0\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%261%5C%5C1%262%5Cend%7Barray%7D%5Cright%5D%20%5C%5CB%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%262%5C%5C2%260%5Cend%7Barray%7D%5Cright%5D)
Then, both are symmetric, but
![AB = \left[\begin{array}{ccc}2&2\\4&2\end{array}\right]](https://tex.z-dn.net/?f=AB%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%5C%5C4%262%5Cend%7Barray%7D%5Cright%5D)
Is not, because (AB)₁₂ = 2 and (AB)₂₁ = 4.
e) This is True. In the ith row and the jth column, since A and B are symmetric, we have
. Therefore, A+B is symmetric.
f) Recall that a matrix C is non singular if and only if Det(C) = 0. Also, det(AB) = Det(A)*Det(B), therefore, if one of them is singular, then there would be a 0 in the product, and as a consequence, AB would have determinant 0, from which we can conclude that AB is singular. On the other hand, if Det(A) and Det(B) are both different from 0, then Det(AB) ≠ 0, so we can conclude that, if both A and B are non singular, then AB also is not singular. As a consequence,
i) AB is non singular
ii) AB is singular
iii) AB is singular
g) False, if
![A = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \\B = \left[\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&-1\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5CB%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%260%260%5C%5C0%26-1%260%5C%5C0%260%26-1%5Cend%7Barray%7D%5Cright%5D)
Then neither A nor B are singular, however the sum is the null matrix, therefore it is singular in spite of being a sum of 2 non singular matrix.