Inverse of matrices

[hitler didi]

HOW TO FIND THE INVERSE of matrice
for eg ::::
(3 4)
(1 2)

this is the matrice but the answer what i get when i do the inverse is not correct the formula is that for the 1st no in 1st coloum you change postion with 2 and ,then you change the signs of 4 & 1 but what will i do if se do the same with the help of determinant of the inverse.

 

[soroban]

Hello, hitler didi!

Find the inverse of: /\(\displaystyle A \:=\:\begin{pmatrix}3&4 \\ 1&2 \end{pmatrix}\)

Inverse of a 2-by-2 matrix: .\(\displaystyle \begin{pmatrix}a&b \\ c&d\end{pmatrix}\)
. . (1) Interchange \(\displaystyle a\) and \(\displaystyle d.\)
. . (2) Change the signs of \(\displaystyle b\) and \(\displaystyle c.\)
. . (3) Divide through by the determinant.


Given: .\(\displaystyle A \:=\:\begin{pmatrix}3&4\\ 1&2\end{pmatrix}\)


Step (1): .\(\displaystyle \begin{pmatrix}2 & . \\ . & 3\end{pmatrix}\)


Step (2): .\(\displaystyle \begin{pmatrix}.&\text{-}4 \\ \text{-}1&. \end{pmatrix}\)


Step (3): .\(\displaystyle D \:=\: (3)(2) - (4)(1) \:=\:2\)

. . . . . . . \(\displaystyle A^{-1} \;=\;\begin{pmatrix}\frac{2}{2} &\text{-}\frac{4}{2} \\ \text{-}\frac{1}{2} & \frac{3}{2}\end{pmatrix} \;=\; \begin{pmatrix}1&\text{-}2 \\ \text{-}\frac{1}{2} & \frac{3}{2} \end{pmatrix}\)

 

[hitler didi]

Hello, hitler didi!


Inverse of a 2-by-2 matrix: .\(\displaystyle \begin{pmatrix}a&b \\ c&d\end{pmatrix}\)
. . (1) Interchange \(\displaystyle a\) and \(\displaystyle d.\)
. . (2) Change the signs of \(\displaystyle b\) and \(\displaystyle c.\)
. . (3) Divide through by the determinant.


Given: .\(\displaystyle A \:=\:\begin{pmatrix}3&4\\ 1&2\end{pmatrix}\)


Step (1): .\(\displaystyle \begin{pmatrix}2 & . \\ . & 3\end{pmatrix}\)


Step (2): .\(\displaystyle \begin{pmatrix}.&\text{-}4 \\ \text{-}1&. \end{pmatrix}\)


Step (3): .\(\displaystyle D \:=\: (3)(2) - (4)(1) \:=\:2\)

. . . . . . . \(\displaystyle A^{-1} \;=\;\begin{pmatrix}\frac{2}{2} &\text{-}\frac{4}{2} \\ \text{-}\frac{1}{2} & \frac{3}{2}\end{pmatrix} \;=\; \begin{pmatrix}1&\text{-}2 \\ \text{-}\frac{1}{2} & \frac{3}{2} \end{pmatrix}\)

oh okay i got it thanks

 

[mmm4444bot]

the answer what i get when i do the inverse is not correct

As always, we cannot see what you are doing. How many times do we have to tell you this?!

what will i do if se do the same with the help of determinant of the inverse

Say again? :-?

You're trying to find the determinant of the inverse matrix?

 

[hitler didi]

sorry i think you people till now have got fed up of my questions right!!!!!!!!!!!!

As always, we cannot see what you are doing. How many times do we have to tell you this?!





Say again? :-?

You're trying to find the determinant of the inverse matrix?

so here is the question:
if i want to find inverse my frist & second step should be as you have done and then take out the detrminent

 

[mmm4444bot]

i think you people till now have got fed up of my questions right!!!!!!!!!!!!

Wrong!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :cool:

How old are you?

You post exercises and say, "My work is not correct".

You don't post your work.

We repeatedly ask you to post your work.

You don't want to.

How can we discuss your work, when you do not post it?

You expect us to read your mind? To come see you, lol?

I think the reason that you continue to ignore us is that you think it is easier to simply post exercises and wait for answers.

I am not fed up with you; I am fed up with having to repeatedly ask you to show your work. :cool:

Why do you not follow our posting guidelines?

Look here: We wrote THIS PAGE especially for you.

(Not everything has to be a revolution.)

 

[Bob Brown MSEE]

HOW TO FIND THE INVERSE of matrice
for eg ::::
(3 4)
(1 2)

Row operations work also....

A =	(3 4)	B =	(1-2)	C =	( 1 0)	D =	(2 0)
	(1 2)		(0 1)		(-1 1)		(0 1)

By applying these matrices....

B*A=	(1 0)	C*B*A=	(1 0)	D*C*B*A=	(2 0)	= 2I
	(1 2)		(0 2)		(0 2)

We see that 2(A^-1)= D*C*B, so

2(A-1)=	(2 0)	*	( 1 0)	*	(1 -2)
	(0 1)		(-1 1)		(0 1)

2(A-1)=	(2 0)	*	( 1 -2)
	(0 1)		(-1 3)

2(A-1) =	( 2 -4)
	(-1 3)

 

[HallsofIvy]

The most fundamental way of doing this is to use the definition of "inverse matrix" directly:
\(\displaystyle \begin{bmatrix}3 & 4 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}3a+ 4c & 3b+ 4d \\ a+ 2c & b+ 2d\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\)

So we have the four equations 3a+ 4c= 1, 3b+ 4d= 0, a+ 2c= 0, and b+ 2d= 1 (actually 2 sets of two equations). From a+ 2c= 0, a= -2c so 3a+ 4c= -6c+ 4c= -2c= 1. c= -1/2 and then a= -2(-1/2)= 1. From 3b+ 4d= 0, b= (4/3)d so b+ 2d= (4/3)d+ 2d= (10/3)d= 1 and d= 3/10. b= (4/3)(3/10)= 4/10= 2/5. That is, the inverse matrix is \(\displaystyle \begin{bmatrix}1 & \frac{2}{5} \\ -\frac{1}{2} & \frac{3}{10}\end{bmatrix}\).