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Yanka [14]
3 years ago
15

Im bord yallllllll ahhh :)

Mathematics
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

You are bord and I'm stuck in a Geometrie assignment because I have a lot of missing work so I have to submit them before June 05 :(

Step-by-step explanation:

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Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of
dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix\mid{A-\lambda I}\mid=0

Where I is identity matrix of the order of given matrix

I=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Substitute the values then, we get

\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0

(1-\lambda)(\lamda^2)-0+0=0

\lambda^2-\lambda^3=0

\lambda^3-\lambda^2=0

Hence, characteristic polynomial =\lambda^3-\lambda^2=0

b.We have to find the eigen value  for given matrix

\lambda^2(1-\lambda)=0

Then , we get \lambda=0,0,1-\lambda=0

\lambda=1

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix A-I

E_1=N(A-I)

A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]

Apply R_1\rightarrow R_1+R_3

A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]

Now,(A-I)x=0[/tex]

Substitute the values then we get

\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0

Then , we get x_3=0

Andx_1-x_2=0

x_1=x_2

Null space N(A-I) consist of vectors

x=\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]

For any scalar x_1

x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]

E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

\left[\begin{array}{ccc}1\\1\\0\end{array}\right]

Eigen space corresponding to 0 eigen value

N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]

(A-0I)x=0

\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0

\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0

Then, x_1+x_3=0

x_1=0

Substitute x_1=0

Then, we get x_3=0

Therefore, the null space consist of vectors

x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

\left[\begin{array}{ccc}0\\1\\0\end{array}\right]

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