Math, asked by reddyarchitha01, 8 months ago

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Find the number of 2 x 2 matrices that can be formed by using 1, 2, 3, 4 wihtout repetion​

Answers

Answered by kasiram569
5

Answer:

A 2x2 matrices has 4 elements a11,a12,a21,a22.

Here given repetition is allowed.

So we can arrange the 1,2,3 and 4 numbers at a11 in 4 ways.

And in the remaining places a12,a21 and a22 also in 4 ways.

So, the numbers combinations formed by using the 1,2,3 and 4 digits when repetitions allowed to from a 2x2 matices=4*4*4*4=256 ways.

Step-by-step explanation:

2x2 Matrix structure shown below:

a11 a12

a21 a22

Now available digits =1,2,3,4

So,we have to construct 2x2 matrices using these digits given that repetition is allowed.

Now,a11 can be occupied by any of 4 digits,

=>a11 can be filled in 4 ways.

Similarily a12,a21,a22 too can be filled un 4 ways.

Hence Total possible combinations of these elements are 4×4×4×4=256.

Hence 256 different matrices can be constructed .

Answered by Anonymous
6

Answer:

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What is the number of 2x2 matrices that can be formed by using 1, 2, 3, and 4 when repetitions are allowed?

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9 Answers

Abhishek Anil Deshmukh, studied Astronomy & Mathematics at IITians PACE

Answered May 16, 2020

2x2 matrices have 2*2 = 4 spaces to fill elements in.

Lets start form top-right:

it can take any number out of the four so it has

top-right: 4 options

Now going to to top-left:

As top right has already used up one of the number from 1, 2, 3, 4. As we cannot repeat that number, we are only left with 3 options for the top-left so,

top-left: 3 options

Moving onto the bottom-right:

The top row took up 2 of the elements from 1, 2, 3, 4. As we cannot repeat any numbers, we are only left with 2 elements for bottom-right so,

bottom-right: 2 options

Now for the last place(bottom left):

All the other places have been filled so we don’t really have any other options than what will be left so,

bottom-left: 1 option

Now we know how many possible options all the places have while not repeating any number, we can just go ahead and multiply them, so

4*3*2*1 = 24 ways to do it.

Note: You can start from any element of the matrix, it doesn’t have to be in any particular order either.

Step-by-step explanation:

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