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Ilya
2 months 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:
Inessa [12.5K]2 months ago
6 0
The question lacks details. Below is the complete version provided.

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 the appropriate dimension.

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

             c_{2}=\frac{-214}{10}

Step-by-step explanation: The identity matrix is a square matrix with 1's along its main diagonal and 0's elsewhere. For example, 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]

When solving the 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]

The multiplication of a matrix by a scalar results in each term being scaled by that scalar. Matrices of different sizes cannot be combined.

So, we structure the equation as follows:

\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 can be written as:

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

Various methods are available to solve this system. One way is to multiply the second equation by -1 and then add the equations together:

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}Following this, substitute variables back into one of the equations to 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}

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

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