Answer:
See picture and explanation below.
Step-by-step explanation:
With this information, the matrix A that you can find is the transformation matrix of T. The matrix A is useful because T(x)=Av for all v in the domain of T.
A is defined as ![T=([T(v_1)] [T(v_2] [T(v_3)])\text{ where }[T(v_i)]](https://tex.z-dn.net/?f=T%3D%28%5BT%28v_1%29%5D%20%5BT%28v_2%5D%20%5BT%28v_3%29%5D%29%5Ctext%7B%20where%20%7D%5BT%28v_i%29%5D)
denotes the vector of coordinates of 
respect to the basis 
(we can apply this definition because 
forms a basis for the domain of T).
The vector of coordinates can be computed in the following way: if 
then ![[T(v_i)]=(a_1,a_2,a_3)^t](https://tex.z-dn.net/?f=%5BT%28v_i%29%5D%3D%28a_1%2Ca_2%2Ca_3%29%5Et)
.
Note that we have all the required information: 
then ![[T(v_1)]=(0,1,0)^t](https://tex.z-dn.net/?f=%5BT%28v_1%29%5D%3D%280%2C1%2C0%29%5Et)
hence ![[T(v_2)]=[T(v_3)]=(1,0,0)^t](https://tex.z-dn.net/?f=%5BT%28v_2%29%5D%3D%5BT%28v_3%29%5D%3D%281%2C0%2C0%29%5Et)
The matrix A is on the picture attached, with the multiplication A(1,1,1).
Finally, to obtain the output required at the end, use the properties of a linear transformation and the outputs given:
In this last case, we can either use the linearity of T or multiply by A.
18/13 or in simplified form, 1 and 5/13.
Here's our equation.
We want to find out when it returns to ground level (h = 0)
To find this out, we can plug in 0 and solve for t.
So the ball will return to the ground at the positive value of
seconds.
What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!
Hi I don’t really understand can you explain a little bit please
