The question is missing parts. Here is the complete question.
Let M = ![\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
. Find 
and 
such that 
, where 
is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.
Answer: 
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
![M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
![M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2Bc_%7B1%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%20%2Bc_%7B2%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6c_%7B1%7D%265c_%7B1%7D%5C%5C-1c_%7B1%7D%26-4c_%7B1%7D%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dc_%7B2%7D%260%5C%5C0%26c_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
And the system of equations is:
There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:
With , substitute in one of the equations and find :
<u>For the equation, </u><u> and </u><u />
Given:
The power generated by an electrical circuit (in watts) as a function of its current x (in amperes) is modeled by
To find:
The current which will produce the maximum power.
Solution:
We have,
Differentiate with respect to x.
...(i)
To find the extreme point equate P'(x)=0.
Divide both sides by -30.
Differentiate (i) with respect to x.
(Maximum)
It means, the given function is maximum at x=4.
Therefore, the current of 4 amperes will produce the maximum power.
Answer:
Step-by-step explanation:
Given the equation as
apply multiplication property of equality where you multiply every term by 5
3x-15=60------------------apply addition property of equality
3x-15+15=60+15
3x=75--------------------------appy division property of equality by dividing both sides by 3
3x/3=75/3
x=25
Answer:
285 boxes are in the display
Step-by-step explanation:
Given data
top layer box = 1
last row box = 81
to find out
how many box
solution
we know that every row is a square so that if the bottom layer has 81 squares it mean this is 9² and every row has one lesser box
so that next row will have 8^2 and than 7² and so on till 1²
so we can say that cubes in the rows as that
Sum of all Squares = 9² + 8² +..........+ 1²
Sum of Squares positive Consecutive Integers formula are
Sum of Squares of Consecutive Integers = (1/6)(n)(n+1)(2n+1)
here n = 9 so equation will be
Sum of Squares of Consecutive Integers = (1/6) × (9) × (9+1) × (2×9+1)
Sum of Squares of Consecutive Integers = 285
so 285 boxes are in the display
