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boyakko [2]
1 year ago
8

Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.

Mathematics
1 answer:
Katyanochek1 [597]1 year ago
6 0

The question is missing parts. Here is the complete question.

Let M = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]. Find c_{1} and c_{2} such that M^{2}+c_{1}M+c_{2}I_{2}=0, where I_{2} is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: c_{1} = \frac{-16}{10}

             c_{2}=\frac{-214}{10}

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

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

M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]

M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]

Solving equation:

\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]

And the system of equations is:

6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)

6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15

10c_{1} = -16

c_{1} = \frac{-16}{10}

With c_{1}, substitute in one of the equations and find c_{2}:

6c_{1}+c_{2}=-31

c_{2}=-31-6(\frac{-16}{10} )

c_{2}=-31+(\frac{96}{10} )

c_{2}=\frac{-310+96}{10}

c_{2}=\frac{-214}{10}

<u>For the equation, </u>c_{1} = \frac{-16}{10}<u> and </u>c_{2}=\frac{-214}{10}<u />

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A formula for 6 gallon of light green paint use 3/8 gallons of white paint. Liam has 9/16 gallon of white paint. Does liam have
FromTheMoon [43]

Answer:

Yes he have enough white paint.

Step-by-step explanation:

We are given

6 gallons of light green paint use → \frac{3}{8} gallons of white paint

using the unitary method

1 gallons of white paint → 6*8/3 gallons of light green paint

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but Liam have 9/16 gallons of white paint

9/16 gallons of white paint → 16* 9/16 gallons of light green paint

                                             → 9 gallons of light green paint

Yes Liam have enough white paint to make 8 gallons of light green paint.  

3 0
2 years ago
A belt and a wallet cost $42 while 7 belts and 4 wallets cost $213. Calculate the cost of each item.
Fittoniya [83]
Let, the cost of a belt = x
cost of a wallet = y

Then, system of equations would be:
x + y = 42
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Multiply 1st equation by 4, 
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Substitute it from 2nd equation, 
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Now, substitute it in 1st equation, 
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In short, Belt costs $15 and wallet costs $27

Hope this helps!
3 0
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Read 2 more answers
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Answer:

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Step-by-step explanation:

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Length = 6 ft

Width = 4 ft

Perimeter = 2 (l+w)

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2nd garden

The length and width are 2 times the 1st garden

Length = 2 *6 = 12

Width = 2 *4 = 8

Perimeter = 2 (l+w)

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Percent change = (new - old )/old * 100 percent

The 1st garden is the old garden = 20  and the 2nd garden is the new garden = 40

Substituting in

Percent change = (40-20)/20 = 20/20 =100 *100 percent

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Answer:

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Step-by-step explanation:

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Checking a probability of transmitting becomes "p".

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= 1-(1-P)^{N}-N[P(1-P)^{N}]

So that the above seems to be the right answer.

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