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I am trying to find the inverse matrix of $$\begin{pmatrix} \ln\left(x\right) & -1\\ \:\:1 & \ln\left(x\right) \end{pmatrix}$$ using the Gauss-Jordan method. Using a different method I could already find that the inverse matrix is:

$$\frac{1}{\ln ^2\left(x\right)+1}\begin{pmatrix}\ln \left(x\right)&-\left(-1\right)\\ -1&\ln \left(x\right)\end{pmatrix}=\begin{pmatrix}\frac{\ln \left(x\right)}{\ln ^2\left(x\right)+1}&\frac{1}{\ln ^2\left(x\right)+1}\\ -\frac{1}{\ln ^2\left(x\right)+1}&\frac{\ln \left(x\right)}{\ln ^2\left(x\right)+1}\end{pmatrix}$$

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  • $\begingroup$ What’s your question? This matrix only has two rows, so there aren’t a lot of steps involved in the process. $\endgroup$ – amd Sep 15 '18 at 1:44
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Suppose that the matrix $A$ is given by

$$ \begin{bmatrix} \ln(x) & -1 \\ 1 & \ln(x) \end{bmatrix} \tag{1} $$

using Gaussian Elimination $$ U=A, L=I \tag{2} $$ $$ \ell_{21} = \frac{u_{21}}{u_{11}} = \frac{1}{\ln(x)} \tag{3} $$ the point is to find the coefficient to zero the column $$ u_{2,1:2} = u_{2,1:2} -\frac{1}{\ln(x)}u_{1,1:2} \tag{4}$$ $$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} -\frac{1}{\ln(x)}\begin{bmatrix} \ln(x) & -1 \end{bmatrix} \tag{5}$$

which gives us

$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{\ln(x)}\end{bmatrix} \tag{6} $$

$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{\ln(x)}\end{bmatrix} \tag{7} $$

$$ u_{2,1:2} = \begin{bmatrix} 0 & \frac{\ln^{2}(x)+1}{\ln(x)} \end{bmatrix} \tag{8} $$

updating the matrix

$$ U = \begin{bmatrix} \ln(x) & -1 \\ 0 & \frac{\ln^{2}(x)+1}{\ln(x)} \end{bmatrix} \tag{9} $$

$$ L = \begin{bmatrix} 1 & 0 \\ \frac{1}{\ln(x)} & 1 \end{bmatrix} \tag{10} $$

$$ A = LU \tag{11} $$

So you'd find $U^{-1}L^{-1}$

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  • $\begingroup$ You’ve “reduced” the problem to finding the inverses of two matrices and then multiplying them together. If you instead carry the row-reduction through the rest of the way, which requires only two more operations, you’ll end up with $A^{-1}$, anyway. $\endgroup$ – amd Sep 15 '18 at 4:29
  • $\begingroup$ I’ve posted no answers because no question was asked. Where have I called anything trivial? Bringing $[A\mid I]$ to RREF form produces $[I\mid A^{-1}]$ directly. An $LU$ decomposition, while interesting, is something of a diversion. $\endgroup$ – amd Sep 15 '18 at 4:41
  • $\begingroup$ $U$ is upper-triangular, but not even in echelon form, let alone in row-reduced echelon form, which for any invertible $2\times2$ matrix is $I$. To quote you, “don’t you see that?” And yes, there is a difference. $\endgroup$ – amd Sep 15 '18 at 4:52

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