How to find the inverse of a given matrix?

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In summary, the problem is trying to find the inverse of a matrix using the A-1:frown: property, but is not able to find the answer.
  • #1
haha1234
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Homework Statement


The problem is posted below.

Homework Equations


In part b,I think I can use E1E2...A=I this property.
But I cannot find the answer.

The Attempt at a Solution


By using the property,I multiply the inverse giving with σ2and σ1.
However,the answer is not correct!
So how to get the correct answer?
Thanks!
 

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My neck hurts. Doesn't it say something in the guidelines about posting pictures ?
 
  • #3
It's not that hard to type in the matrix itself. Also, if you know the basic definitions, you should not have much difficulty deciding what elementary matrices are used. For example, we are given the matrix,
[tex]A_1= \begin{bmatrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 0 & -1 & -3 \end{bmatrix}[/tex]
and are asked what elementary matrix, [itex]\sigma_1[/itex] converts that to
[tex]A_2= \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & -1 & -3 \end{bmatrix}[/tex]

There are three kinds of row operations:
1) multiply every number in a single row by a number
2) swap two rows
3) add a multiple of one row to another
Comparing [itex]A_1[/itex] and [itex]A_2[/itex], we see that rows 1 and 2 have not changed so we have not swapped two rows. The second row has changed from [2 5 4] to [0 1 2]. If we were multiplying by a number, to go from "2" to "0" we would have to multiply by 0 but then we would have "0" in all positions so the row operation was not multiplying by a number. And since the first number in the last row is 0, adding any multiple of that row to the second row would not have changed that number. So the new second row must be [2 5 4] plus a multiple of [1 2 1]. To go from 2 to 0, we must have subtracted 2 which is the same as adding -2 times 1. Adding -2 times 2 to 5 gives 1 and adding -2 times 1 to 4 gives 2. That is exactly what we wanted. So the row operation must be "add -2 times each number in the first row to the corresponding number in the second row".

To get the elementary matrix that gives that, do that row operation to the identity matrix. Adding -2 times each member in the first row to the corresponding number in the second row of the identity matrix gives
[tex]\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]
That is the elementary matrix [itex]\sigma_1[/itex].

Now, continue like that
 
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  • #4
HallsofIvy said:
It's not that hard to type in the matrix itself. Also, if you know the basic definitions, you should not have much difficulty deciding what elementary matrices are used. For example, we are given the matrix,
[tex]A_1= \begin{bmatrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 0 & -1 & -3 \end{bmatrix}[/tex]
and are asked what elementary matrix, [itex]\sigma_1[/itex] converts that to
[tex]A_2= \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & -1 & -3 \end{bmatrix}[/tex]

There are three kinds of row operations:
1) multiply every number in a single row by a number
2) swap two rows
3) add a multiple of one row to another
Comparing [itex]A_1[/itex] and [itex]A_2[/itex], we see that rows 1 and 2 have not changed so we have not swapped two rows. The second row has changed from [2 5 4] to [0 1 2]. If we were multiplying by a number, to go from "2" to "0" we would have to multiply by 0 but then we would have "0" in all positions so the row operation was not multiplying by a number. And since the first number in the last row is 0, adding any multiple of that row to the second row would not have changed that number. So the new second row must be [2 5 4] plus a multiple of [1 2 1]. To go from 2 to 0, we must have subtracted 2 which is the same as adding -2 times 1. Adding -2 times 2 to 5 gives 1 and adding -2 times 1 to 4 gives 2. That is exactly what we wanted. So the row operation must be "add -2 times each number in the first row to the corresponding number in the second row".

To get the elementary matrix that gives that, do that row operation to the identity matrix. Adding -2 times each member in the first row to the corresponding number in the second row of the identity matrix gives
[tex]\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]
That is the elementary matrix [itex]\sigma_1[/itex].

Now, continue like that
...I know how to find the corresponding elementary matrix,but I don't know how to find the inverse using the A-1:frown:
 

FAQ: How to find the inverse of a given matrix?

What is the definition of an inverse?

An inverse is a mathematical operation that reverses the effect of another operation. For example, the inverse of addition is subtraction, and the inverse of multiplication is division.

How do you find the inverse of a function?

To find the inverse of a function, you must first solve for the independent variable (x). Then, switch the positions of the independent and dependent variables to create the inverse function. For example, the inverse of f(x) = 2x + 3 would be written as f^-1(x) = (x-3)/2.

What is the difference between a function and its inverse?

A function maps one input value to one output value, while its inverse maps one output value to one input value. In other words, the input and output variables are reversed in the inverse function.

Are there any restrictions when finding an inverse?

Yes, there are a few restrictions when finding an inverse. The function must be one-to-one, meaning each input value has a unique output value. The function must also be onto, meaning every output value has a corresponding input value. Additionally, the function must be continuous and have a unique inverse for each point on its graph.

Why is finding the inverse important in mathematics?

Finding the inverse of a function is important because it allows us to solve equations and perform operations that would otherwise be impossible. Inverses are also useful in real-world applications, such as finding the original amount from a percentage increase or decrease.

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