Thread: Inverse of a matrix?

1. Inverse of a matrix?

Hi, I'm having a problem finding the inverse of a matrix in that I have no idea where I am going wrong. I am given two simultaneous equations and told to find the inverse of A and therefore determine the value of x.

$\displaystyle 2x_1 + 3x_2 = 1\ and\ x_1 + 7x_2 = 6$. I have put this in the form Ax = b as asked by the question.

$\displaystyle \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right] = \left[\begin{array}{c}1\\6\end{array}\right]$.

I know that if I multiply both sides by the inverse I get $\displaystyle x = A^{-1}b$. To find the inverse from here $\displaystyle A^{-1} = \frac{1}{det(A)}\cdot A_{cof}T$ (where the T is for transposed).

So $\displaystyle A = \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \ \ A_{cof} = \left[\begin{array}{cc}2&-1\\-3&7\end{array}\right] \ \ A_{cof}T = \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$

I calculated the det(A) as being 11 (14 -3) so $\displaystyle A^{-1} = \frac{1}{11} \cdot \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$

If something hasn't already gone wrong then here is where it doesn't make sense from me. Just to make sure that it was correct I decided to multiply A by the inverse I just got and to see if I got the identity matrix. I didn't. Can anyone help me in finding where I have gone wrong. I can do the rest from there

Thanks

2. Originally Posted by Beard
Hi, I'm having a problem finding the inverse of a matrix in that I have no idea where I am going wrong. I am given two simultaneous equations and told to find the inverse of A and therefore determine the value of x.

$\displaystyle 2x_1 + 3x_2 = 1\ and\ x_1 + 7x_2 = 6$. I have put this in the form Ax = b as asked by the question.

$\displaystyle \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right] = \left[\begin{array}{c}1\\6\end{array}\right]$.

I know that if I multiply both sides by the inverse I get $\displaystyle x = A^{-1}b$. To find the inverse from here $\displaystyle A^{-1} = \frac{1}{det(A)}\cdot A_{cof}T$ (where the T is for transposed).

So $\displaystyle A = \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \ \ A_{cof} = \left[\begin{array}{cc}2&-1\\-3&7\end{array}\right] \ \ A_{cof}T = \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$

I calculated the det(A) as being 11 (14 -3) so $\displaystyle A^{-1} = \frac{1}{11} \cdot \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$

If something hasn't already gone wrong then here is where it doesn't make sense from me. Just to make sure that it was correct I decided to multiply A by the inverse I just got and to see if I got the identity matrix. I didn't. Can anyone help me in finding where I have gone wrong. I can do the rest from there

Thanks
Your error is in finding $\displaystyle A_{cof}T$, which should be::

$\displaystyle \left[\begin{array}{cc}7&-3\\-1&2\end{array}\right]$

3. Hello, Beard!

Your inverse matrix is wrong . . .

given: . $\displaystyle \begin{array}{ccc} 2x + 3y &=& 1 \\ x + 7y &=& 6 \end{array}$

I have put this in the form Ax = b as asked by the question.

. . $\displaystyle \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x \\ y\end{array}\right] \:=\: \left[\begin{array}{c}1\\6\end{array}\right]$

So: . $\displaystyle A^{-1} \:=\: \frac{1}{11} \cdot \left[\begin{array}{cc}2&\text{-}3\\\text{-}1&7\end{array}\right]$ . . . . no

There is a "formula" for the inverse of a 2-by-2 matrix.

We have: .$\displaystyle \left[\begin{array}{cc} a&b \\ c&d \end{array}\right]$

[1] Switch the elements on the main diagonal: .$\displaystyle \left[\begin{array}{cc}{\color{red}d}&b\\c&{\color{red}a }\end{array}\right]$

[2] Change the signs on the other diagonal: .$\displaystyle \left[\begin{array}{cc}d & {\color{red}\text{- }}\!b \\ {\color{red}\text{- }}\!c & a\end{array}\right]$

[3] Divide by the determinant: . $\displaystyle {\color{red}\frac{1}{ad-bc}}\,\left[\begin{array}{cc} d & \text{- }\!b \\ \text{- }\!c & a\end{array}\right]$

Therefore: . $\displaystyle \left[\begin{array}{cc}a&b\\c&d\end{array}\right]^{-1} \;=\;\frac{1}{ad-bc}\,\left[\begin{array}{cc}d&\text{- }\!b \\ \text{- }\!c & a \end{array}\right]$

4. I know can see how I went wrong, the examples I was trying to follow had ones for a and c looking at sorobans method so I was unable to distinguish.

Thanks