Matrices, how to get this equality

[Ry122]

How do I get the right side of the equation to equal the left side?$$<br /> \begin{bmatrix}<br /> 1 & 0 \\ <br /> 0 & 1<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> a & 1 \\ <br /> 1 & 1<br /> \end{bmatrix}<br />$$

I tried multiplying the first row by 0 to get rid of a is this is the right first step?

 

[Dick]

They aren't equal because 0 is not equal to 1. Face it. Massaging it won't change that.

 

[Ry122]

Sorry I should have explained a bit better.

What I'm trying to do is find the inverse matrix of
ax + y = 1
x + y = b

I don't think that equal sign should be in there

 

[Dick]

Sorry I should have explained a bit better.

What I'm trying to do is find the inverse matrix of
ax + y = 1
x + y = b

I don't think that equal sign should be in there

Then multiplying the first row by 0 seems like a pretty poor idea. Now you've got NOTHING in the first row. How about multiplying by 1/a, so you can subtract the first row from the second and get a 0 in the lower left part of the original matrix?

 

[HallsofIvy]

So your question is "How do I find the inverse to
$$\left[\begin{array}{cc}a & 1 \\ 1 & 1 \end{array}\right]$$?

One way to do that is to use, directly, the definitiion: Find x, y, z u, so that
$$\left[\begin{array}{cc}a & 1 \\ 1 & 1 \end{array}\right][left[\begin{array}{cc}x & y \\ z & u\end{array}\right]= \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$$.

That is the same as solving the equation ax+ z= 1, ay+ u= 0, x+ z= 0, y+ u= 1 for x, y, z, u.

But I suspect you are talking about "row reduction". The "row operations" are
1) Swap two rows
2) Multiply one row by a number
3) Add a multiple of one row to another.

In order to row reduce [itex]\left[\begin{array}{cc}a & 1 \\ 1 & 1 \end{array}\right] do the following:<br /> <br /> Normally, the simplest way to get a "1" on the diagonal is to multiply the entire row by 1 over whatever is on the diagona. Here, since we already have a '1' directly below a, I would swap the first and second rows to get<br /> $$\left[\begin{array}{cc} 1 & 1 \\ a & 1\end{array}\right]$$<br /> To get a "0" below that, subtract a times the first row from the second (do you see how I decided to use "a time"? Since we have a "1" just above the"a", we use a/1= a.)<br /> $$\left[\begin{array}{cc} 1 & 1 \\ 0& 1-a\end{array}\right]$$<br /> Now you want to get a "1" on the diagonal, in the lower right. Since that is 1- a now, the obvious thing to do is divide that row by 1- a. Notice that does not change the first row:<br /> $$\left[\begin{array}{cc} 1 & 1 \\ 0& 1\end{array}\right]$$<br /> Finally, subtract the second row from the first and you are done:<br /> $$\left[\begin{array}{cc} 1 & 0 \\ 0& 1\end{array}\right]$$<br /> Now if you apply those same row operations to the identity matrix, in the same order, you get the inverse matrix. Of course, it is simpler to write the matrices A and the identity matrix side-by-side and do the row operations to both matrices at the same time.[/itex]

 

[Ry122]

thanks