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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:

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

bryan.blackbee
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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: http://www.mathsisfun.com/algebra/matrix-inverse-row-operations-gauss-jordan.html – Julien 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. – bryan.blackbee Feb 14 '13 at 12:18
  • Following https://en.m.wikipedia.org/wiki/Dual_basis#:~:text=From%20Wikipedia%2C%20the%20free%20encyclopedia,%E2%88%97%20form%20a%20biorthogonal%20system. The dual basis is simply the inverse of the matrix column vectors and in the 3d case the dual vectors are given by :$e^i=\frac{1}{e_i\cdot e_j\times e_k}e_j\times e_k$ where ijk are cyclically passing through 123 – QuantumPotatoïd May 09 '21 at 12:36

3 Answers3

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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.

Ross Millikan
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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. – bryan.blackbee Feb 14 '13 at 12:41
  • btw, the answer is already posted. – bryan.blackbee Feb 14 '13 at 12:41
  • @bryansis2010: It can be quite mechanical. Just do the operations that make the zeros you need appear. Sometimes clever choices can get you more zeros than you are due, as when I subtracted twice row 1 from row 3 and in the approach I suggest in my previous comment. But if you don't see that in a particular case, you can always get them one at a time. – Ross Millikan Feb 14 '13 at 12:45
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$$\mathbf{A}^{-1} = \begin{bmatrix} a & b & c\\ d & e & f \\ g & h & k\\ \end{bmatrix}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} \, A & \, B & \,C \\ \, D & \, E & \,F \\ \, G & \,H & \, K\\ \end{bmatrix}^T = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} \, A & \, D & \,G \\ \, B & \, E & \,H \\ \, C & \,F & \, K\\ \end{bmatrix}$$ If the determinant is non-zero, the matrix is invertible, with the elements of the above matrix on the right side given by $$\begin{matrix} A = (ek-fh) & D = (ch-bk) & G = (bf-ce) \\ B = (fg-dk) & E = (ak-cg) & H = (cd-af) \\ C = (dh-eg) & F = (gb-ah) & K = (ae-bd). \\ \end{matrix}$$

superAnnoyingUser
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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.

MyUserIsThis
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