# Thread: Reduce the matrix [A|I]

1. ## Reduce the matrix [A|I]

Hi,

I need guidance with the following question:

Question: Reduce the matrix [A|I] in order to determine if A is invertible. Compute A^-1 when A is invertible and check that AA^-1 = I. Verify that the (1,1)-entry in the product A^-1 is 1.

A=
2 1
5 3

For starters,I was going to find the determinant of A like this: |A|= ad-bc, but realized the question is asking me to reduce the matrix [A|I]. So how do I reduce [A|I]? Do I multiply matrix A * Identity matrix, then use reduced row echelon like this:

2 1|1 0
5 3|0 1

2. Perform the necessary row operations on the entire matrix $\displaystyle [A|I]$ necessary to take $\displaystyle A$ to $\displaystyle I.$ If you can do it, you've got an invertible matrix $\displaystyle A,$ and you should end up with

$\displaystyle [A|I]\to[I|A^{-1}].$

Make sense?

3. Do you know what it means to "reduce" a matrix? You want to go from that matrix to the identity matrix by "row operations" and there are three kinds of row operations:
1) Divide (or multiply) an entire row by a number
2) Add (or subtract) a multiple of a row to (or from) another row
3) Swap two rows.

Since you want a "1" in the upper left, a good way to start is to divide the entire first row by the number that is in that position- here that is a "2". Then, because you want a "0" below that, subtract the new first row times the number you want to get rid of, here a "5", from the second row.

(Be careful with your terminology. You are not "multiplying" A by the identity matrix- you are simply writing them next to one another so that you can apply the same row operations to both matrices at the same time.)

4. Thanks for your replies Ackbeet and HallsofIvy!

Ok, here is what I've done so far:

2 1 | 1 0 R1 * 1/2
5 3 | 0 1

1 1/2 | 1/2 0
5 3 | 0 1 R2 -5R1

1 1/2 | 1/2 0
0 1/2 | 5/2 1 R2 *2

1 1/2 | 1/2 0 R1 -1/2R2
0 1 | 5 2

1 0 | -2 -1
0 1 | 5 2

Now that I've reduced A to 1s and 0s, how do I know that A is invertible? I noticed that my Identity matrix has changed close to what A is (except for the minus 2 and minus 1 and the 2)

5. So your candidate for $\displaystyle A^{-1}$ is now

$\displaystyle A^{-1}\overset{?}{=}\begin{bmatrix}-2 &-1\\5&2\end{bmatrix}.$

So, multiply it out to check that

$\displaystyle AA^{-1}=A^{-1}A=I.$

Does it work? I think you'll find that you made a sign error in there somewhere in your row reduction.

6. Ackbeet, you were right about the sign:

2 1 | 1 0 R1 * 1/2
5 3 | 0 1

1 1/2 | 1/2 0
5 3 | 0 1 R2 -5R1

1 1/2 | 1/2 0
0 1/2 | -5/2 1 R2 *2

1 1/2 | 1/2 0 R1 -1/2R2
0 1 | -5 2

1 0 | 3 -1
0 1 | -5 2

7. You multiply that out, and I think you'll see that you now have the correct inverse.

Verify that the (1,1)-entry in the product A^-1 is 1.
This is a quote from the OP. I think you meant that the (1,1) entry in the product AA^(-1) is 1? If so, you can see that that's true.

8. Yes you are right Ackbeet!

A^-1 =
3 -1
-5 2

2 1 | 3 -1
5 3 | -5 2

6 -5 -2 + 2
15-15 -5 + 6

1 0
0 1

And yes, I did make a mistake in the question. It should read:
...Verify that the (1,1)-entry in the product AA^-1 is 1.
But how do I verify that the (1,1)-entry in the product AA^-1 is 1? Do I calculate it like this: ad-cb, that is, 1*1 - 0*0 = 1?

Thanks a lot Ackbeet!

9. The (1,1) entry is the upper left corner entry of the matrix you calculated as

$\displaystyle \begin{bmatrix}6-5&-2+2\\15-15 &-5+6\end{bmatrix}.$

So what would you say its value is?

10. Originally Posted by Ackbeet
The (1,1) entry is the upper left corner entry of the matrix you calculated as

$\displaystyle \begin{bmatrix}6-5&-2+2\\15-15 &-5+6\end{bmatrix}.$

So what would you say its value is?
I would say the value of the upper left corner is 6-5 = 1 Thanks a lot Ackbeet!

11. Yep, I'd say you've got it. You're welcome, and have a good one!