# Inverse $T$ matrix of a 3*3 matrix. [closed]

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.

• What has $T$ to do with $A$? Mar 26 at 15:17
• it is A=TDT^(-1), where D=diag(λ1,λ2,λ3) and T=[V1|V2|v3] Mar 26 at 15:23
• @Irini In that case your matrix $T$ of eigenvectors is incorrect! Mar 26 at 15:49
• @Irini: Shouldn't you have $$T = \left( \begin{array}{ccc} 1 & 4 & 23 \\ 0 & 1 & 20 \\ 0 & 0 & 12 \\ \end{array} \right)$$
– Moo
Mar 26 at 16:11
• 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. Mar 26 at 16:14

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

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

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