How to find the inverse of a given matrix?

[haha1234]

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!

 

[BvU]

My neck hurts. Doesn't it say something in the guidelines about posting pictures ?

 

[HallsofIvy]

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,
A_1= \begin{bmatrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 0 & -1 & -3 \end{bmatrix}
and are asked what elementary matrix, \sigma_1 converts that to
A_2= \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & -1 & -3 \end{bmatrix}

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 A_1 and A_2, 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
\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
That is the elementary matrix \sigma_1.

Now, continue like that

 

[haha1234]

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,
A_1= \begin{bmatrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 0 & -1 & -3 \end{bmatrix}
and are asked what elementary matrix, \sigma_1 converts that to
A_2= \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & -1 & -3 \end{bmatrix}

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 A_1 and A_2, 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
\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
That is the elementary matrix \sigma_1.

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