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Marta_Voda [28]
2 years ago
9

If A is a 2 × 2 matrix, then A × I = and I × A =

Mathematics
1 answer:
krok68 [10]2 years ago
3 0

Since the multiplication between two matrices is not <em>commutative</em>, then \vec A\, \times\,\vec I \ne \vec I \,\times \,\vec A, regardless of the dimensions of \vec A.

<h3>Is the product of two matrices commutative?</h3>

In linear algebra, we define the product of two matrices as follows:

\vec C = \vec A \,\times \vec B, where \vec A \in \mathbb{R}_{m\times p}, \vec B \in \mathbb{R}_{p\times n} and \vec C \in \mathbb{R}_{m \times n}     (1)

Where each element of the matrix is equal to the following dot product:

c_{ij} = \left[\begin{array}{cccc}a_{i1}&a_{i2}&\ldots&a_{ip}\end{array}\right]\,\bullet\,\left[\begin{array}{ccc}b_{1j}\\b_{2j}\\\vdots\\b_{pj}\end{array}\right], where 1 ≤ i ≤ m and 1 ≤ j ≤ n.     (2)

Because of (2), we can infer that the product of two matrices, no matter what dimensions each matrix may have, is not <em>commutative</em> because of the nature and characteristics of the definition itself, which implies operating on a row of the <em>former</em> matrix and a column of the <em>latter</em> matrix.

Such <em>"arbitrariness"</em> means that <em>resulting</em> value for c_{ij} will be different if the order between \vec A and \vec B is changed and even the dimensions of \vec C may be different. Therefore, the proposition is false.

To learn more on matrices: brainly.com/question/9967572

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Solve the solution of a system of linear equation<br><br> 1/2 x + y = - 6<br> 6 x + 2 y = 8
Lilit [14]

Answer:

(4,-8)

Step-by-step explanation:

Plug in the answers for the equations.

4 0
3 years ago
An urn contains 2 red marbles and 3 blue marbles. 1. One person takes two marbles at random from the urn and does not replace th
Ghella [55]

Answer:

A) The best way to picture this problem is with a probability tree, with two steps.

The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.

The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.

If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.

If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.

B) So a person can have a red marble and a blue marble in two ways:

1) Picking the red first and the blue last

2) Picking the blue first and the red last

C) P(R&B) = 3/5 = 60%

Step-by-step explanation:

C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5

3 0
3 years ago
Give the first ten terms of the following sequences. You can assume that the sequences start with an index of 1. Logs are to bas
Vitek1552 [10]

Answer:

a)

a1 = log(1) = 0 (2⁰ = 1)

a2 = log(2) = 1 (2¹ = 2)

a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849

a4 = log(4) = 2 (2² = 4)

a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322

a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)

a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807

a8 = log(8) = 3 (2³ = 8)

a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )

a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322

b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:

log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that

a1 = 1

a2 = 2

a3 = 3

a4 = 4

a5 = 5

a6 = 6

a7 = 7

a8 = 8

a9 = 9

a10 = 10

I hope this works for you!!

3 0
3 years ago
i need help on math
dolphi86 [110]
X=2
2x-3=-x+3

add three to both sides
add 1x to each side
should be left with 3x=6
divide both sides by 3 to get x alone
your answer then is x=2
7 0
3 years ago
Which expression is equivalent to b^-3/ab^-3?
Mrrafil [7]
1/a

The b^-3 in the denominator and numerator cancel out
7 0
3 years ago
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