If A and B are equal:
Matrix A must be a diagonal matrix: FALSE.
We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:
![A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]](https://tex.z-dn.net/?f=A%3DB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C4%265%5C%5C7%268%5Cend%7Barray%7D%5Cright%5D)
Both matrices must be square: FALSE.
We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works
Both matrices must be the same size: TRUE
If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.
For any value of i, j; aij = bij: TRUE
Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.
Answer:
It's where they meet! I can't really see the coordinate clearly, but if I could I would tell you already.
( 3 1 ) 2 +3 2 =left parenthesis, start fraction, 1, divided by, 3, end fraction, right parenthesis, squared, plus, 3, squared
Fynjy0 [20]
Answer:
I guess that we have the equation:
(1/3)^2 + 3^2 =
And we want to solve it.
Here just remember that:
3^2 = 3*3 = 9
then:
(1/3)^2 = (1/3)*(1/3) = 1/9
replacing these in the expression, we get:
(1/3)^2 + 3^2 = 1/9 + 9
If we want to write this as a single fraction, we can rewrite:
1/9 + 9 = 1/9 + (9/9)*9
= 1/9 + 81/9 = (1 + 81)/9 = 82/9
Answer: 2
Step-by-step explanation:
2.5(6x-4)=10+4(1.5+0.5x) [Work out the brackets]
15x-10=0+6+2x [Solve]
15x-10=16+2x [Convey the terms]
15x-2x=16+10 [Add up the same terms and calculate]
13x=26 [Divide both parts]
x=2
