Answer:

Step-by-step explanation:
It is a result that a matrix
is orthogonally diagonalizable if and only if
is a symmetric matrix. According with the data you provided the matrix should be

We know that its eigenvalues are
, where
has multiplicity two.
So if we calculate the corresponding eigenspaces for each eigenvalue we have
,
.
With this in mind we can form the matrices
that diagonalizes the matrix
so.

and

Observe that the rows of
are the eigenvectors corresponding to the eigen values.
Now you only need to normalize each row of
dividing by its norm, as a row vector.
The matrix you have to obtain is the matrix shown below
Answer:
x = –8
Step-by-step explanation:
The numerator is 5 (and the denominator is 10).
Answer:
See explanation
Step-by-step explanation:
An equation that represents the path of a diver jumping off a diving board is

The diver jumped to the water when x = 0.
Then

The jumper reached the water when y = 0, then

So, point where the jumper reached the water is
