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3 years ago

Please answers please

Mathematics

1 answer:

Allisa [31]3 years ago

3 0

Answer:

i honestly dont know

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 $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

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

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

5 0

3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPP plzzzzzzzzzzzz example i do anything just help

Snowcat [4.5K]

Answer:

-9/20

(Decimal: -0.45)

Step-by-step explanation: 1/2 -1/4

-2/5

= 1/-5 -1/4

=-1/5- 1/4

= -1/5 +-1/4

= -9/20

5 0

3 years ago

(-1, 3) (-5, -5) (0, -4) (1, 2) (4, 4) (3, 8)

velikii [3]

What is the question

4 0

3 years ago

Complete the rectangle above. What are the coordinates of the missing vertex?

vagabundo [1.1K]

To create the rectangle, you will need the ordered pair (7,3). The first number on the ordered pair is where the point is positioned on the x-axis. Think of it as the horizontal number line. For the second number in the ordered pair you should think of where the point is located on the y-axis. Think of this as the vertical number line.

Go over 7 (first), and then go up 3 (second) to plot the last point.

8 0

3 years ago

For what values of x is x2-36=5x true? 0 -9 and 4 -4 and 9 4 and -9 9 and 4

sveticcg [70]

Answer:

-4 and 9

Step-by-step explanation:

x2-36=5x'

x*x - 5x - 36 = 0

(x - 9) (x + 4) = 0

roots 9 and -4

8 0

4 years ago