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I am given a $3 \times3$ matrix and am asked to find the inverse using elementary row operations. I know how they work, but have no idea of which steps to apply first, followed by which steps.

First, the matrices:

$$\begin{pmatrix} 1 & 1 & -3\\ 2 & 1 & -3\\ 2 & 2 & 1 \end{pmatrix}$$

All I know thus far is that, if there is a series of operations (pre-multipliers)

$E_nE_{n-1}...E_2E_1A$ that reduces to the identity matrix, the same sequence $ E_nE_{n-1}...E_2E_1I$ reduces to the inverse of $A$, $A^{-1}$.

Any help? If not, I will use another method already because this is not working thus far.

UPDATE

Thanks to the community, I got the final answer:

$$\begin{pmatrix} -1 & 1 & 0\\ \frac8 7 & -1 & \frac 3 7\\ \frac{-2}{7} & 0 & \frac 1 7 \end{pmatrix}$$

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In the first position of the first row you have number 1. What multiple of the first row you need to subtract from the second row so that after this operation the second row has zero in the first position? Can you obtain zero in the first position of the third row? –  Martin Sleziak Feb 14 '13 at 12:09
    
Here is what you can do, for instance: mathsisfun.com/algebra/… –  1015 Feb 14 '13 at 12:11
    
You might want to look up "Gauss–Jordan elimination". That is an algorithm that does exactly this. –  CBenni Feb 14 '13 at 12:13
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Honestly, I really did not know that was called Gauss-Jordan elimination. I'll go find out what that actually means. Damn my lecture notes are so vague...this is so much more comprehensive...I'll try it and then finish it. –  bryansis2010 Feb 14 '13 at 12:18

3 Answers 3

up vote 5 down vote accepted

I was taught to augment the matrix with the identity, then apply the row operations:

$$\begin{pmatrix} 1 & 1 & -3 & 1 & 0 & 0\\ 2 & 1 & -3&0&1&0\\ 2 & 2 & 1&0&0&1 \end{pmatrix}$$

Subtract twice row 1 from row 2 and twice row 1 from row 3 (yes, this is two operations)

$$\begin{pmatrix} 1 & 1 & -3 & 1 & 0 & 0\\ 0 & -1 & 3&-2&1&0\\ 0 & 0 & 7&-2&0&1 \end{pmatrix}$$

Multiply row 2 by -1 and row 3 by $\frac 17$

$$\begin{pmatrix} 1 & 1 & -3 & 1 & 0 & 0\\ 0 & 1 & -3&2&-1&0\\ 0 & 0 & 1&-2/7&0&1/7 \end{pmatrix}$$

Subtract row 2 from row 1, then add three times the third to the second and you are there. The right three columns will be your inverse.

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For the record, this is usually called Gauss–Jordan elimination. –  CBenni Feb 14 '13 at 12:13
    
I bet Ross, and in fact almost all of us up to and including the OP, knew that, @CBenni . The important fact here is that the answer above shows how to actually carry on the process described by the OP in a doable and easy way –  DonAntonio Feb 14 '13 at 12:15
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@DonAntonio I do not think the OP knew. There is very little reason discussing it. However if the OP didnt know, he can look it up for further information. –  CBenni Feb 14 '13 at 12:17
    
@bryansis2010: there are many ways to get there. I now see I could have subtracted row 1 from row 2 to get zeros in the second and third columns, then swapped row 1 and row 2. Maybe that is an easier approach. –  Ross Millikan Feb 14 '13 at 12:27
    
i have solved it, though! probably I will continue to practise because these questions are really hard..."seeing" the solution takes time. –  bryansis2010 Feb 14 '13 at 12:41

That way of calculating ineverses (which was very slow imo), was to write it this way:

$$\begin{pmatrix}1 & 1 & -3 & | & 1 & 0 & 0\\ 2 & 1 & -3 & | & 0 & 1 & 0\\ 2 & 2 & 1 & | & 0 & 0 & 1 \end{pmatrix}$$

Ten make those elementary operations on the first one to reduce to the identity, while making the same operations on the right one. When you get the identity on the left one, what you have on the right one will be the inverse.

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There is no algortihm that is faster for abritrary matrices. –  CBenni Feb 14 '13 at 12:15
    
@CBenni Doing it by hand, I find it a lot faster doing it by the determinant method. –  MyUserIsThis Feb 14 '13 at 12:31
    
That does not work for matrices of sizes above 3x3 iirc? I agree it can be faster for small matrices however, eventhough the number of computations is higher. –  CBenni Feb 14 '13 at 12:57
    
@CBenni I think it does work for abritrarily big matrices... According to wikipedia, it does. –  MyUserIsThis Feb 14 '13 at 13:56
    
yes, but then you have to use the adjungated matrix, not the transposed; This makes it alot more work. –  CBenni Feb 14 '13 at 16:18

$$\mathbf{A}^{-1} = \begin{bmatrix} a & b & c\\ d & e & f \\ g & h & k\\ \end{bmatrix}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} \, A & \, B & \,C \\ \, D & \, E & \,F \\ \, G & \,H & \, K\\ \end{bmatrix}^T = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} \, A & \, D & \,G \\ \, B & \, E & \,H \\ \, C & \,F & \, K\\ \end{bmatrix}$$ If the determinant is non-zero, the matrix is invertible, with the elements of the above matrix on the right side given by $$\begin{matrix} A = (ek-fh) & D = (ch-bk) & G = (bf-ce) \\ B = (fg-dk) & E = (ak-cg) & H = (cd-af) \\ C = (dh-eg) & F = (gb-ah) & K = (ae-bd). \\ \end{matrix}$$

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The question is about finding the inverse of the matrix. –  1015 Feb 14 '13 at 12:12
    
@julien Fixed it. –  Student Feb 14 '13 at 12:22
    
Edit Does not make difference. Read Original Post "am asked to find the inverse using elementary row operations" –  user45099 Feb 14 '13 at 12:22
    
Or use another method already. –  Student Feb 14 '13 at 12:24
    
Yes, great. But I think the OP really wants to do this by elementary row operations. +1 anyway. –  1015 Feb 14 '13 at 12:25

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