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This should be fairly simple as I know matrix A can be found by $$A = E_1^{-1} E_2^{-1}...E_k^{-1}$$ So it should go that $$A^-$$ can be found by $$E_kE_(k-1)...E_1$$

But for some reason my numbers aren't matching up with the book.

I have a matrix $$\begin{bmatrix}1&0&-1\\0 &6&-1\\0&0&4\end{bmatrix}$$

I find that I can get an Identity Matrix from this matrix by doing (1/6)R2 -> R2, (1/4)R3 -> R3, 1/6R3 + R2 -> R2, R3 + R1 -> R1. From there I can find the inverse of the elementary matrices no problem but for some reason my normal E does not multiply into the inverse. Did I do something wrong in my steps? It should be noted I need to solve the matrix using elementary matrices.

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  • $\begingroup$ Also can anyone tell me how to properly make _ and ^ work for more than just the next thing I type? I'm having trouble figuring out why it doesn't work as seen in my examples above; I am sorry for the inconvenience/eyesore. $\endgroup$
    – Charlatan
    Sep 15, 2018 at 12:00
  • $\begingroup$ Use "{}" to wrap all of your $-1$, e.g. $A^{-1}$ vs $A^-1$. Same for "E_{k-1}" as in $E_{k-1}$ v.s. $E_k-1$. $\endgroup$
    – xbh
    Sep 15, 2018 at 12:08
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    $\begingroup$ It’s very hard to say what you did wrong if you don’t show the details of your steps. It might be something as simple as multiplying the elementary matrices together in the wrong order. $\endgroup$
    – amd
    Sep 15, 2018 at 19:40
  • $\begingroup$ Is... Is that really a worry? If I multiply the E's in wrong order can that really ruin the answer? $\endgroup$
    – Charlatan
    Sep 15, 2018 at 21:28

1 Answer 1

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If $E_1\cdots E_n A = I$, then $A^{-1} = E_1\cdots E_n I$.

So you can transform $A$ into $I$ and simultaneously transform $I$ to $A^{-1}$ using the same transformations:

$$ \left[\begin{array}{rrr|rrr} 1&0&-1 & 1 & 0 & 0\\0 &6&-1& 0 & 1 & 0\\0&0&4& 0 & 0& 1 \end{array}\right] \sim \left[\begin{array}{rrr|rrr} 1&0&-1 & 1 & 0 & 0\\0 &1&-\frac16& 0 & \frac16 & 0\\0&0&1& 0 & 0& \frac14 \end{array}\right]\sim \left[\begin{array}{rrr|rrr} 1&0&0 & 1 & 0 & \frac14\\0 &1&0& 0 & \frac16 & \frac1{24}\\0&0&1& 0 & 0& \frac14 \end{array}\right]$$

so $A^{-1} = \begin{bmatrix} 1 & 0 & \frac14 \\ 0 & \frac16 & \frac1{24} \\ 0 & 0 & \frac14\end{bmatrix}$.


Using elementary matrices with this notation, our transformations are $$L_{3,2}\left(\frac16\right)L_{3,1}(1)D_3\left(\frac14\right)D_2\left(\frac16\right)A = I$$

so \begin{align} A^{-1} &= L_{3,2}\left(\frac16\right)L_{3,1}(1)D_3\left(\frac14\right)D_2\left(\frac16\right)\\ &= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & \frac16 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac14 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & \frac16 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & \frac14 \\ 0 & \frac16 & \frac1{24} \\ 0 & 0 & \frac14\end{bmatrix} \end{align}

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  • $\begingroup$ Forgot I could do this; thank you for reminding me of the augmented matrices role $\endgroup$
    – Charlatan
    Sep 15, 2018 at 21:27
  • $\begingroup$ just remembered why I didn't do the augmented matrix method; I was suppose to solve by using elementary matrices. Do you know how to do that @mechanodroid? $\endgroup$
    – Charlatan
    Sep 16, 2018 at 1:03
  • $\begingroup$ @Charlatan Sure, have a look now. $\endgroup$ Sep 16, 2018 at 7:39
  • $\begingroup$ Not sure how to read this notation; would you mind explaining what D and L mean? I have the same values you got for the elementary matrices I noticed but I ordered mine backwards from yours; am I suppose to put $$E_1$$ as the first transformation? $\endgroup$
    – Charlatan
    Sep 18, 2018 at 5:47
  • $\begingroup$ @Charlatan The notation is explained in the linked page. $D_i(\lambda)$ is a matrix which multiplies row $i$ by a scalar $\lambda$. $L_{i,j}(\lambda)$ is a matrix which adds row $i$ multiplied by $\lambda$ to row $j$. $\endgroup$ Sep 18, 2018 at 8:25

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