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It is given that, $A$ is $3 \times 3$ matrix and that $$A\begin{pmatrix} x\\y\\z\\\end{pmatrix}= 5\begin{pmatrix} p\\q\\r\\\end{pmatrix}$$ and that $$x= 3p+2q-4r \\ y=p+5q-2r \\ z=7p-6q+3r \\$$ then how can I find $A^\text{-1}$ ?

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  • $\begingroup$ solve it for $p,q,r$ $\endgroup$ Aug 11, 2017 at 17:25
  • $\begingroup$ multiply both sides of equation by $A^{-1}/5$, and fill in the standard basis vectors for $(p,q,r)$. This will give the columns of $A^{-1}$. Note the inverse exist, since $A$ is surjective by the equation. $\endgroup$
    – M. Van
    Aug 11, 2017 at 17:31

1 Answer 1

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Hint: $$\begin{bmatrix}x\\y\\z\end{bmatrix} = 5A^{-1}\begin{bmatrix}p\\q\\r\end{bmatrix}.$$ The first row of $5 A^{-1} = \begin{bmatrix}3 & 2 & -4\end{bmatrix}.$

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  • $\begingroup$ please give me more hints... $\endgroup$
    – raf
    Aug 11, 2017 at 17:56
  • $\begingroup$ @MubtasimFuad I gave you the first row of $5A^{-1}$. Compare it with the first equation $x = 3 p + 2q - 4r$. Can you figure out how I got $[3~2~-4]$ from the equation? For the second row, use the second equation. Likewise, the third row can be obtained from the third equation. $\endgroup$
    – Math Lover
    Aug 11, 2017 at 18:00
  • $\begingroup$ Thank you, I think I got it. so $$A^{-1}=\begin{pmatrix} 3/5 & 2/5 & -4/5\\ 1/5 & 1 & -2/5\\ 7/5 & -6/5 & 3/5\\ \end{pmatrix}$$ ? $\endgroup$
    – raf
    Aug 11, 2017 at 18:16
  • $\begingroup$ I'm glad that you have obtained the correct value of $A^{-1}$. $\endgroup$
    – Math Lover
    Aug 11, 2017 at 18:17

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