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3 years ago

PLEASE HELP ME ASAP! Giving brainliest SHOW YOUR WORK FOR BOTH QUESTIONS

Mathematics

1 answer:

EleoNora [17]3 years ago

3 0

Answer:

The Proof for question no 20 and 21 are below.

Step-by-step explanation:

20.

Given:

∠PTR≅∠RSP

PT ≅ RS

To Prove:

ΔPQT ≅ ΔRQS

Proof:

In ΔPQT and ΔRQS

Statements Reasons

1. ∠PTR ≅ ∠RSP 1 .Given

2. ∠PQT ≅ ∠ROS 2. Vertical Opposite Angle Theorem.

3. PT ≅ RS 3. Given

4. ΔPQT ≅ ΔRQS 4. By A-A-S Congruence Postulate ....Proved

21.

Given:

PO ≅ SO

O is the Mid Point of NT

To Prove:

∠N ≅ ∠T

Proof:

In ΔPON and ΔSOT

Statements Reasons

1. PO ≅ S O 1 .Given

2. ∠PON ≅ ∠SOT 2. Vertical Opposite Angle Theorem.

3. NO ≅ TO 3. O is the Mid Point of NT

4. ΔPON ≅ ΔSOT 4. By S-A-S Congruence Postulate

5. ∠N ≅ ∠T 5. Corresponding Parts of Congruent Triangles are Congruent .( CPCTC )...Proved

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A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei

Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

$A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]$

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Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

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which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are $\lambda=4,3,-1$

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression $A-\lambda I$ and solve the system $(A-\lambda I)x =0$ by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider $\lambda = 4$ . We get the matrix

$\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]$

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

$(x,y,z) = (5z,5z,z) = z(5,5,1)$

So, the base for this eigenspace is the vector (5,5,1).

If $\lambda = 3$ then the base is (0,4,3) and if $\lambda = -1$ then the base is (0,0,1)

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3 years ago

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Answer:

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Step-by-step explanation:

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I hope this is correct.

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Answer:

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