Could you give me your feedback ?

Determine the inverse of the following Matrix: $$A = \begin{bmatrix} 1 & 2 & -1 \\ -3 & 1 & 2 \\ -2 & 2 & 1 \end{bmatrix}$$

We want to perform row operations until we get the identity matrix (we could also do it with the determinant method). We will apply the same operations to an identity matrix and the resulting matrix will be the inverse of $A$.

We add 3 times first row to second row, 2 times first row to third row. Then we add -1 third row to second row. Then we add -2 second row to first row and -6 second row to third row. Then we multiply third row by -1. Then we add 1 third row to first row. And we get our identity matrix.

If we apply those operations to an identity matrix, we get

$$A^{-1} = \begin{bmatrix} 3 & 4 & -5 \\ 1 & 1 & -1 \\ 4 & 6 & -7 \end{bmatrix}$$

Is this correct ?

Thanks for your help !

  • 1
    $\begingroup$ You simply need to check that $A(A^{-1})$ is indeed the identity matrix. That's what the inverse of a matrix means. What is the confusion? $\endgroup$ May 6 at 17:38
  • $\begingroup$ @user1046533 There isn't a confusion, was just wondering if my procedure is correct In fact multiplying both matrices gives the identity matrix $\endgroup$
    – wengen
    May 6 at 17:44

2 Answers 2


I can not write comment because of the reputation so I wanted to write in here. You can always check that your calculation is True or not $ A. A^{-1} = I_n = A^{-1}.A$


Yes, your answer is correct. You can check by doing a dot product of the first row of $A$ with the first column of $B$ and it is $1$. That's usually good enough to know whether you got it right or wrong.


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