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I don't know how to do this question, I appreciate if anyone can help. Thank you.

(a) For what values of $a,b$, and $c$ is the following matrix symmetric? $$\begin{bmatrix} -3 & 6a-c & 6a+2b\\ a & 2 & 4\\ a+7b & c & a \end{bmatrix}$$ (b) An $n\times n$ matrix $A$ is called skew-symmetric if $A^T = -A$. What values of $a,b, c,$ and $d$ now make the following matrix skew-symmetric? $$\begin{bmatrix} d & 6a-c & 6a+2b\\ a & 0 & 4-5d\\ a+7b &c &0 \end{bmatrix}$$

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  • $\begingroup$ Please use MathJax. $\endgroup$ – Em. Feb 7 '17 at 4:32
  • $\begingroup$ @ Max I would use mathJax but I'm not familiar with coding so forgive me about this issue. Thank you $\endgroup$ – Armando Feb 7 '17 at 6:16
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A symmetric matrix $A$ means that all entries $A_{ij}$ are symmetric respect to the diagonal, i.e., $A_{ij}=A_{ji}$ for all $i$, $j$.

In your case, you need $6a-c=a$, $6a+2b=a+7b$, and $c=4$ for the first matrix to be symmetric. From this, you can solve for $a,b,c$.

In the second case, the definition of skew-symmetric can be written as that $A_{ij}=-A_{ji}$, so you can similarly equate each term to its equivalent in the transposed neagtive matrix.

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  • $\begingroup$ Thanks for your help. Can you please tell me more regarding part B? do I have C=-(4-5d)? then what would be value of C? I got little bit confuse $\endgroup$ – Armando Feb 7 '17 at 6:22
  • $\begingroup$ Yes, $c=-(4-5d)$, and also $a=-(6a-c)$ and so on. With this, you'll set up a system of four equations and for unknowns that you should be able to solve easily. $\endgroup$ – Anna SdTC Feb 7 '17 at 6:48
  • $\begingroup$ What is fourth equation? d is -1? $\endgroup$ – Armando Feb 7 '17 at 7:17
  • $\begingroup$ No, if you compute $A^T$ you get $d$ as the first element, and $-A$ has $-d$ as first element, so $-d=d$. $\endgroup$ – Anna SdTC Feb 7 '17 at 7:37
  • $\begingroup$ so when you say -d = d.- means d = d? what that negative sign do after d? ( sorry if I ask basic or funny questions) $\endgroup$ – Armando Feb 7 '17 at 18:04

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