To solve:
(x-10)(x-10)
Answer: x^2-20x+100
Answer:
![\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The <em>transpose of a matrix </em>
is one where you swap the column and row index for every entry of some original matrix 
. Let's go through our first matrix row by row and swap the indices to construct this new matrix. Note that entries with the same index for row and column will stay fixed. Here I'll use the notation 
and 
to refer to the entry in the i-th row and the j-th column of the matrices 
and 
respectively:
Constructing the matrix from those entries gives us
![P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=P%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
which is option a. from the list.
Another interesting quality of the transpose is that we can geometrically represent it as a reflection over the line traced out by all of the entries where the row and column index are equal. In this example, reflecting over the line traced from 2 to 1 gives us our transpose. For another example of this, see the attached image!
Answer:
Step-by-step explanation:
we have to find the dot product here.
From the formula of dot product of vectors,
we know-
if X=<a,b,c> and Y=<d,e,f>
Then dot product of X and Y is-
X.Y=ad+be+cf
Here, we get
v=<2 , -8 , -8>
and w=<-2 , 6 , -5>
So, the dot product is-
v . w=2(-2)+(-8)(6)+(-8)(-5)
=-4-48+40
=-52+40
=-12
So,
v . w= -12
