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I have the matrix $$ A= \begin{bmatrix} 1 & 4 & 1\\ 0 & 2 & 5\\ 0 & 0 & 5 \end{bmatrix}. $$ I have found the $$ T= \begin{bmatrix} 1 & 4 & 1\\ 0 & 0 & 1\\ 0 & 1 & 3/5 \end{bmatrix}. $$ How can I find the $T^{-1}$? I already know the $\frac{1}{|T|}$ part but I am confused with the adjugate matrix.

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    $\begingroup$ What has $T$ to do with $A$? $\endgroup$ Commented Mar 26 at 15:17
  • $\begingroup$ it is A=TDT^(-1), where D=diag(λ1,λ2,λ3) and T=[V1|V2|v3] $\endgroup$
    – Irini
    Commented Mar 26 at 15:23
  • $\begingroup$ @Irini In that case your matrix $T$ of eigenvectors is incorrect! $\endgroup$ Commented Mar 26 at 15:49
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    $\begingroup$ @Irini: Shouldn't you have $$T = \left( \begin{array}{ccc} 1 & 4 & 23 \\ 0 & 1 & 20 \\ 0 & 0 & 12 \\ \end{array} \right)$$ $\endgroup$
    – Moo
    Commented Mar 26 at 16:11
  • $\begingroup$ Can you clarify what part of the definition of adjugate are you struggling with? It's the transpose of the matrix whose $ij$-entry is $\det M_{ij}$, where $M_{ij}$ is the determinant of the minor, where row $i$ and column $j$ have been removed. In other words, the number $\det M_{ij}$ is placed in the $ji$-entry of the adjugate. $\endgroup$ Commented Mar 26 at 16:14

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I think the easiest way (other than asking Wolfram Alpha) is to do a row echelon reduction on the augmented matrix

$$ U = \left[ \begin{array}{ccc|ccc} 1 & 4 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 3/5 & 0 & 0 & 1 \\ \end{array} \right] $$

This gives

$$ V = \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 7/5 & \mbox{-}4 \\ 0 & 1 & 0 & 0 & \mbox{-}3/5 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ \end{array} \right] $$

which means the inverse is

$$ T^{-1} = \left[ \begin{array}{ccc} 1 & 7/5 & \mbox{-}4 \\ 0 & \mbox{-}3/5 & 1 \\ 0 & 1 & 0 \\ \end{array} \right]$$

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    $\begingroup$ Unfortunately, OP's matrix of eigenvectors is already incorrect... Moo's comment to the post gives the correct matrix $T$. $\endgroup$ Commented Mar 26 at 16:12

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