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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solve it for $p,q,r$ – Dr. Sonnhard Graubner Aug 11 '17 at 17:25
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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. – M. Van Aug 11 '17 at 17:31
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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 $5 A^{-1} = \begin{bmatrix}3 & 2 & -4\end{bmatrix}.$
Math Lover
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@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. – Math Lover Aug 11 '17 at 18:00
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Thank you, I think I got it. so $$A^{-1}=\begin{pmatrix} 3/5 & 2/5 & -4/5\ 1/5 & 1 & -2/5\ 7/5 & -6/5 & 3/5\ \end{pmatrix}$$ ? – raf Aug 11 '17 at 18:16
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