Question 1174343: Q−1: [5×2 marks] Answer each of the following as True or False justifying your answers:
If A and B are n×n matrices, then 〖(A+B)〗^2=A^2+2AB+B^2.
If A is an n×n invertible symmetric matrix, then (A^(-1) )^k is symmetric, for any positive integer k.
Let A be an n×n matrix. If A is invertible, then |A|≠0.
If A and B are n×n matrices, then |A+B^T |=|A^T+B|.
The vectors (a,0,0), (1,a,0) and (2,3,a) in R^3 are linearly independent for all a∈R.
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
I'll do the first two problems to get you started.
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Problem 1
(A+B)^2 = (A+B)*(A+B)
(A+B)^2 = C*(A+B) ............ let C = A+B
(A+B)^2 = C*A+C*B ............ Distribute
(A+B)^2 = (A+B)*A+(A+B)*B ......... plug in C = A+B
(A+B)^2 = A*A+B*A+A*B+B*B ...... distribute again; be careful about the order
(A+B)^2 = A^2+B*A+A*B+B^2
We stop here because B*A is not the same as A*B. So we cannot combine them to get 2AB or 2BA. Matrix multiplication is not commutative in general (there are some cases where it works, but overall it doesn't work).
Therefore, (A+B)^2 = A^2+2AB+B^2 is false when A,B are two nxn matrices.
It's only true when AB = BA.
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Problem 2
The transpose operation of a matrix swaps the rows and columns. For instance, row 1 turns into column 1, and vice versa.
With matrix notation, the T indicates transpose.
Writing A^T means "transpose of matrix A"
One useful property to take advantage of here is that
(A*B)^T = B^T*A^T
I won't prove that here, but these links are a good resource for that
https://math.stackexchange.com/questions/3208939/transpose-of-product-of-matrices
https://math.stackexchange.com/questions/1440305/how-to-prove-abt-bt-at
Matrix A is symmetric if and only if A = A^T
Square both sides
A = A^T
A^2 = (A^T)^2
A^2 = (A^T)*(A^T)
A^2 = (A*A)^T .......... apply the transpose property mentioned
A^2 = (A^2)^T
This proves that if A is symmetric, then so is A^2
This example can be extended to show that if A is symmetric, then so is A^3, and so on.
By induction, if A is symmetric, then A^k is symmetric for any positive integer k.
If A is invertible, then A^(-1) multiplies with it to lead to the identity matrix I
A*A^(-1) = A^(-1)*A = I
Let's focus on A*A^(-1) = I
Apply the transpose operation to both sides
A*A^(-1) = I
[A*A^(-1)]^T = I^T
[A^(-1)]^T*A^T = I
we can see that for the matrix A^T, the matrix [A^(-1)]^T plays the role of the inverse.
Put another way, let B = [A^(-1)]^T
We can see that
B*A^T = I
which leads to
B*A^T = I
(B*A^T)*(A^T)^(-1) = I*(A^T)^(-1)
B * [A^T*(A^T)^(-1)] = (A^T)^(-1)
B * I = (A^T)^(-1)
B = (A^T)^(-1)
Therefore,
[A^(-1)]^T = (A^T)^(-1)
A very similar situation occurs with A^T*B = I, but we'll be using left-multiplication to arrive at the same equation shown above.
What I've been building up to so far is to show that- If matrix A is symmetric, then A^k is symmetric for any positive integer k
- If matrix A is invertible, then [A^(-1)]^T = (A^T)^(-1)
Such properties have probably been discussed and/or proven in your textbook at this point. If not, then likely they've been left as a homework exercise.
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Let's get back to the problem at hand. We know matrix A is invertible. This means [A^(-1)]^T = (A^T)^(-1) by that second bullet point.
But we also know matrix A is symmetric. So A = A^T, which leads to
[A^(-1)]^T = (A^T)^(-1)
[A^(-1)]^T = A^(-1)
A^(-1) = [A^(-1)]^T
which shows that the inverse matrix A^(-1) is symmetric, but only if A is symmetric.
If A is symmetric, then so is A^(-1)
But if A^(-1) is symmetric, we can use the first property mentioned in the bullet points above to conclude that ( A^(-1) )^k is also symmetric for any positive integer k.
Therefore, the statement that matrix A being symmetric and invertible leads to ( A^(-1) )^k also being symmetric, for any positive integer k, is a true statement.
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