Answer: Matrix B is non- invertible.
Step-by-step explanation:
A matrix is said to be be singular is its determinant is zero,
We know that if a matrix is singular then it is not invertible. (1)
Or if a matrix is invertible then it should be non-singular matrix. (2)
Given : A and B are n x n matrices from which A is invertible.
Then A must be non-singular matrix. ( from 2 )
If AB is singular.
Then either A is singular or B is singular but A is a non-singular matrix.
Then , matrix B should be a singular matrix. ( from 2 )
So Matrix B is non- invertible. ( from 1 )
If tan(<span>θ) is negative, then </span><span>θ must be either in Q-II or else in Q-IV.
Fortunately, the question tells us that it's in Q-II.
If you draw a circle on the x- and y-axes, then draw a right triangle
in Q-II, then mark the legs 3 and -2, then the hypotenuse of the
triangle ... also the radius of the circle ... is √13 .
Look for the angle whose tangent is -3/2.
tangent = (opposite) / (adjacent)
So the side opp</span>osite is the 3 and the side adjacent is the -2.
For that same angle, cosine = (adjacent) / (hypotenuse) .
The adjacent side is still the -2, and the hypotenuse is √13 .
So the cosine of the same angle is
- 2 / √13 .
To rationalize the denominator (get that square root out of there),
multiply top and bottom by √13 . Then you have
(- 2 / √13) · (√13 / √13)
= - 2 √13 / 13 .
Answer:
1445
Step-by-step explanation:
2601*(9/9-4/9)
The answer is c sorry if i am wrong
Answer:
Please see below.
Explanation:
The coordinates of point C are (a+b,c).
The coordinates of the midpoint of diagonal ¯¯¯¯¯¯AC are (a+b2,c2).
The coordinates of the midpoint of diagonal ¯¯¯¯¯¯BD are (a+b2,c2).
¯¯¯¯¯¯AC and ¯¯¯¯¯¯BD intersect at point E with coordinates (a+b2,c2).
By the definition of midpoint, ¯¯¯¯¯¯AE≅¯¯¯¯¯¯CE and ¯¯¯¯¯¯BE≅¯¯¯¯¯¯DE.
Therefore diagonals ¯¯¯¯¯¯AC and ¯¯¯¯¯¯BD bisect each other.
