Given matrix A = [[a b][c d]], the inverse is supposed to be

K

kryms3n

New member
Joined
Dec 4, 2006
Messages
14
I am really confused how to work with inverse matrices. I have to show how this matrix is an inverse of A:

Code: 
A= [a   b]
   [c   d]
I know that the inverse is supposed to be:

Code: 
(1/ ad -bc) [d   -b]
            [-c   a]
But how? Help, please! Thank you!
 
There is really nothing to show there. You have all the 'parts'.

\(\displaystyle \L\left[ {\begin{array}{cc}
a & b \\
c & d \\
\end{array}} \right],\quad \& \left( {ad - bc} \right) \not= 0\quad \Rightarrow \left[ {\begin{array}{cc}
a & b \\
c & d \\
\end{array}} \right]^{ - 1} = \left[ {\begin{array}{cc}
{\frac{d}{{ad - bc}}} & {\frac{{ - b}}{{ad - bc}}} \\
{\frac{{ - c}}{{ad - bc}}} & {\frac{a}{{ad - bc}}} \\
\end{array}} \right]\)
 
Re: CHALLENGE PROBLEMS need help

Hello, kryms3n!

i am really confused how to do inverse
. . There's not much to "do" here.

I have to show that \(\displaystyle \L A \:=\:\begin{bmatrix} a & b\\c & d\end{bmatrix}\)

and \(\displaystyle \L A^{-1} \:=\:\frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\)
Click to expand...

How do we know if two matrices are inverses of each other?
. . . . . Their product is the identity matrix.

So multiply \(\displaystyle A\times\,A^{^{-1}}\) or \(\displaystyle A^{^{-1}}\,\times\,A\) and see if you get: \(\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\)

 
I have to show how I got from the first equation to the second equation, using the three elementary row operations: interchanging two rows, multiplying a row, or adding a row (or a multiple of a row) to another row.

These are the teacher's instructions, and we don't get points if we don't use these rules, and show how we used them. I'm really confused on how to do it.
 
Well o.k. You have a great deal of messy work to do.
Solve the system of 4 equations for w,x,y & z.

\(\displaystyle \L \begin{array}{l}
\left[ {\begin{array}{cc}
a & b \\
c & d \\
\end{array}} \right]\left[ {\begin{array}{cc}
w & x \\
y & z \\
\end{array}} \right] = \left[ {\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}} \right] \\
\left[ {\begin{array}{cc}
{aw + by} & {ax + bz} \\
{cw + dy} & {cx + dz} \\
\end{array}} \right] = \left[ {\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}} \right] \\
\left\{ {\begin{array}{c}
{aw + by = 1} \\
{ax + bz = 0} \\
{cw + dy = 0} \\
{cx + dz = 1} \\
\end{array}} \right. \\
\end{array}\)
 
alright

alright
i think i explained the answer wrong

I have to prove that the first equation
has the inverse of the second equation
by using the three elementary row operations
interchange
multiply
and add
 
kryms3n said:
I have to prove that the first equation
has the inverse of the second equation...
Click to expand...
You haven't posted equations, and I'm not sure what "the inverse of an equation" would mean. Do you perhaps mean that you need to show that the first matrix is the inverse of the other...? If so, what method(s) did your book and/or class give you for finding inverse matrices?

Thank you.

Eliz.
 
You must log in or register to reply here.
Top