Question

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.

Answer 1

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 $c_{2}$:

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

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

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

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

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

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