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Ket [755]
3 years ago
9

Free p0ints first come first serve

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer:

Tyty :)

Step-by-step explanation:

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Find a fundamental matrix solution for the system x 0 1 7x1 + 4x2 + 12x3, x 0 2 x1 + 2x2 + x3, x 0 3 −3x1 − 2x2 − 5x3. Then find
liq [111]

Here is the correct format for the question.

x'_1 = 7x_1 +4x_2+ 12x_3 , \ \ x'_2 = x_1 + 2x_2 + x_3 ,  \ \ x'_3 = -3x_1 -2x_2 -5x_3 . Then find the solution that satisfies  x \limits ^{\to} = \left[\begin{array}{c}0\\1\\-2\end{array}\right]

Answer:

\mathbf{c_1 = -3, c_2 = 2, c_3 = 2}

Step-by-step explanation:

From the figures given above:

the matrix can be computed as,

\left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right] x \limits ^{\to} = \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = x \limits ^{\to} = \left[\begin{array}{c}x_1'\\x_2'\\x_3'\end{array}\right]

The first thing we need to carry out is to determine the eigenvalues of A,

where:

A = \left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right]

|A-rI|=0

\begin {vmatrix} 7-r&4&12\\1&2-r&1\\-3&-2&-5-r  \end {vmatrix}=0

the eigenvalues are r = 0, 1, 3

However, the eigenvector correlated to each eigenvalue can be calculated as follows.

suppose  r = 0

(A - rI) x = 0

\left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right] \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right]

now the eigenvector is \left[\begin{array}{c}-4\\1\\2\end{array}\right]

However, for eigenvalue = 1, we have :  \left[\begin{array}{c}-4\\1\\2\end{array}\right]

for eigenvalue = 3, we have:\left[\begin{array}{c}-2\\-1\\1\end{array}\right]

The solution now can be computed as :

x(t)=  c_1 \left[\begin{array}{c}-4\\1\\2\end{array}\right] + c_2e^t \left[\begin{array}{c}-4\\3\\1\end{array}\right]+ c_3e^{3t} \left[\begin{array}{c}-2\\-1\\1\end{array}\right]

Similarly, the fundamental matrix solution is:

\left[\begin{array}{ccc}-4&-4e^t&-2e^{3t}\\1&3e^t&-e^{3t}\\2&e^t&e^{3t}\end{array}\right]

-4c_1 -4c_2-2c_3 =0 \\ \\ c_1 + 3c_2 -c_3 = 1\\ \\ 2c_1+c_2 +c_3 = -2

Solving the above equation, we get:

\mathbf{c_1 = -3, c_2 = 2, c_3 = 2}

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6 0
3 years ago
In what order should you perform the operations in the expression 4×3−12÷2+5?
ioda

Answer:

Bidmas(brakets-indices-division-multiplication-addtion-subtraction) therfore

4*3-6+5=12-6+5=12-1=11

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3 years ago
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