Answer:
Step-by-step explanation:
7 + (-5)= 2
Answer:
Entries of I^k are are also identity elements.
Step-by-step explanation:
a) For the 2×2 identity matrix I, show that I² =I
![I^{2}=\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \times \left[\begin{array}{cc}1&0\\0&1\end{array}\right] \\\\=\left[\begin{array}{cc}1\times 1+0\times 0&1\times 0+0\times 1\\0\times 1+1\times 0&0\times 0+1\times1\end{array}\right] \\\\=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%5Ctimes%201%2B0%5Ctimes%200%261%5Ctimes%200%2B0%5Ctimes%201%5C%5C0%5Ctimes%201%2B1%5Ctimes%200%260%5Ctimes%200%2B1%5Ctimes1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Hence proved I² =I
b) For the n×n identity matrix I, show that I² =I
n×n identity matrix is as shown in figure
Elements of identity matrix are

As square of 1 is equal to 1 so for n×n identity matrix I, I² =I
(c) what do you think the enteries of Ik are?
As mentioned above

Any power of 1 is equal to 1 so kth power of 1 is also 1. According to this Ik=I
Answer:
Step-by-step explanation:
The best way to do this is to use your LCM and eliminate the fractions. To find the LCM you have to use all the denominators as a multiplier so the denominator in each term cancels out. We will first factor the x-squared term to simplify and see what 2 factors are hidden there.
factors to (x + 2)(x - 2). That means that our 3 denominators that make up our LCM are x(x+2)(x-2). We will mulitply that in to each term in our rational equation, canceling out the denominators where applicable.
![x(x+2)(x-2)[\frac{2}{(x-2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}]](https://tex.z-dn.net/?f=x%28x%2B2%29%28x-2%29%5B%5Cfrac%7B2%7D%7B%28x-2%29%7D%2B%5Cfrac%7B7%7D%7B%28x-2%29%28x%2B2%29%7D%3D%5Cfrac%7B5%7D%7Bx%7D%5D)
In the first term, the (x-2) will cancel leaving us with
x(x+2)[2] which simplifies to
![x^2+2x[2]](https://tex.z-dn.net/?f=x%5E2%2B2x%5B2%5D)
In the second term, the (x+2)(x-2) cancels out leaving us with
x[7].
In the last term, the x cancels out leaving us with
(x+2)(x-2)[5] which simplifies to
![x^2-4[5]](https://tex.z-dn.net/?f=x%5E2-4%5B5%5D)
Now we will distribute through each cancellation:
2x²+4x;
7x;
5x²-20
Putting them all together we have
2x² + 4x + 7x = 5x² - 20
Combining like terms gives us a quadratic:
3x² - 11x - 20 = 0
Factor that however you find it easiest to factor quadratics and get that
x = 5 and x = -4/3
You should ask your parent/guardian
or teacher if needed! Merry Christmas!