What is the constant of the polynomial 2x3 - 8x2 + 3x - 7
1 answer:
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Answer:
A(10,-2).
Step-by-step explanation:
It is given that, on a coordinate plane, a triangle has points A'(-2, 10), B'(-6, 4), and C'(-10, 8).
We need to find the pre-image of vertex A' if the image shown on the graph was created by a reflection across the line y = x.
If a figure reflected across the line y=x, then
It is given that A'(-2,10).
On comparing both sides, we get
Therefore, the coordinates of preimage of A' are (10,-2).
Answer:
the P matrix you are looking for is P=(1/) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]]
Step-by-step explanation:
Answer:
For an orthogonal diagonalization of any matrix you have to:
1º) Find the matrix eigenvalues in a set order.
2º) Find the eigenvectors of each respective eigenvalues.
Tip: You can write the matrix A like A = D P
3º) D is the diagonal matrix with each eigenvalue (in order) in the diagonal.
4º) Write P as the normalized eigenvectors in order (in columns).
Tip 2: Remember, ·P = I, so if A =
D P, then:
P A = P
D P
= I D I = D
So the P we are looking for is the of the diagonalization.
Tip 3: In this case, A is a block matrix with null nondiagonal submatrixes, therefore its eigenvalues can be calculated by using the diagonal submatrixes. The problem is reduced to calculate the eigenvalues of A₁₁ = A ₂₂ = [[5 3],[3 5]]
Solving:
1º)the eigenvalues of A₁₁ are {8,2}, therefore the D matrix is
2º) the eigenvectors of A₁₁ are P₈= {[1 1]T} P₂= {[1 -1]T}, therefore normalizing the eigenvectors you obtain P = (1/) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]] (you can see that P =
in this case).
As said in "tip 2": P A = P
D P
= I D I = D
So the P obtained is the one you are looking for.