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This question is in my assignment. We are not allowed to use any symbol to represent any elementary row and column operations used in the solution. We must solve it step-by-step. Please help me to check my solution word by word including my spelling and grammar.

Question:

Find the inverse of

$$A=\begin{pmatrix}2& 2& 3\\ 2& 5& 3\\ 1& 0& 8\end{pmatrix}$$

by using only elementary row operations.

Solution:

We begin by forming the matrix $\begin{pmatrix} A & | & I_3 \end{pmatrix}=\left(\begin{array}{ccc|ccc}2 & 2 & 3 & 1 & 0 & 0\\2 & 5 & 3 & 0 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\end{array}\right)$. Interchanging the first and third rows of the matrix $\begin{pmatrix} A & | & I_3 \end{pmatrix}$, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\2 & 5 & 3 & 0 & 1 & 0\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$. Adding $(-2)$ times the first row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\2 & 5 & 3 & 0 & 1 & 0\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$ to its second row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 5 & -13 & 0 & 1 & -2\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$. Multiplying the second row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 5 & -13 & 0 & 1 & -2\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$ by $\frac{1}{5}$, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$. Adding $(-2)$ times the first row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$ to its third row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 2 & -13 & 1 & 0 & -2\end{array}\right)$. Adding $(-2)$ times the second row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 2 & -13 & 1 & 0 & -2\end{array}\right)$ to its third row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & -\frac{39}{5} & 1 & -\frac{2}{5} & -\frac{6}{5}\end{array}\right)$. Multiplying the third row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & -\frac{39}{5} & 1 & -\frac{2}{5} & -\frac{6}{5}\end{array}\right)$ by $(-\frac{5}{39})$, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$. Adding $(\frac{13}{5})$ times the third row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$ to its second row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$. Adding $(-8)$ times the third row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$ to its first row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{40}{39} & -\frac{16}{39} & -\frac{3}{13}\\0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$. Thus, $A^{-1}=\begin{pmatrix}\frac{40}{39} & -\frac{16}{39} & -\frac{3}{13}\\ -\frac{1}{3} & \frac{1}{3} & 0\\ -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{pmatrix}$.

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  • $\begingroup$ ...and the question is...? $\endgroup$
    – DonAntonio
    Nov 17, 2013 at 18:28
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    $\begingroup$ you can check your own solutions: Multiply the original $A$ but the $A^{-1}$ you found. If you did everything right you should get the identity matrix. $\endgroup$ Nov 17, 2013 at 18:28
  • $\begingroup$ bluebit.gr/matrix-calculator/calculate.aspx $\endgroup$
    – Zhoe
    Nov 17, 2013 at 18:29
  • $\begingroup$ Kudos for writing all the pmatrices. I can't see that you have missed anything. I am unable to check your arithmetic at the moment, but personally I can't see how you should be expected to do more steps. $\endgroup$
    – Arthur
    Nov 17, 2013 at 18:34
  • $\begingroup$ Yup...your solution is correct.. $\endgroup$
    – user93470
    Nov 18, 2013 at 8:36

3 Answers 3

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Bingo!

Well done. I even checked by multplying $AA^{-1}$ and $A^{-1}A$ and obtained $I_3$ each time. And this is confirmed here, as well.

Your solution is correct, very well-written, and easy to follow (albeit there's a lot to follow!).

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Your solution is correct.$ $

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This all looks fine to me. I'm not sure how much more you expect from an answer. As some advice, you might like to, at the end, actually multiply the two matrices just to make sure they are mutually inverse.

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