Right-inverse of matrix

[TsAmE]

Find all the right-inverses of the matrix \(\displaystyle \begin{pmatrix}1 & -2 &3 \\ -4 & 5 & -6\end{pmatrix}\)

Attempt:

\(\displaystyle \textbf{AR} = \mathbf{I_2}\)

\(\displaystyle \begin{pmatrix}1 & -2 &3 \\ -4& 5 &-6 \end{pmatrix}\begin{pmatrix}x1 &y1 \\ x2&y2 \\ x3&y3 \end{pmatrix} = \begin{pmatrix}1 & 0\\ 0& 1\end{pmatrix}\)

\(\displaystyle \begin{pmatrix}1 & -2 & 3 & | 1 & 0\\ -4& 5 &-6 & |0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2 & 3 & | 1 & 0\\ 0& -3 &6 &|4 & 1\end{pmatrix}\) Row2New = Row2 + 4Row1

I am suppose to be getting the same identity matrix:\(\displaystyle \mathbf{I_2}\) on the LHS of the 2nd part of the gauss reduction, but I am not

 

[galactus]

I deleted my previous post because I noticed you were looking for all right inverses.

\(\displaystyle \begin{bmatrix}1&-2&3\\-4&5&-6\end{bmatrix}\cdot \begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\).................[1]

By using \(\displaystyle A^{T}\left(A\cdot A^{T}\right)^{-1}\), we get \(\displaystyle \Large \begin{bmatrix}\frac{-17}{18}&\frac{-4}{9}\\ \frac{1}{9}&\frac{1}{9}\\ \frac{13}{18}&\frac{2}{9}\end{bmatrix}\)..............[2]

From [1], we get:

\(\displaystyle a-2c+3e=1\)

\(\displaystyle b-2d+3f=0\)

\(\displaystyle -4a+5c-6e=0\)

\(\displaystyle -4b+5d-6f=1\)

Using rref, this becomes:

\(\displaystyle a-e=\frac{-5}{3}\)

\(\displaystyle b-f=\frac{-2}{3}\)

\(\displaystyle c-2e=\frac{-4}{3}\)

\(\displaystyle d-2f=\frac{-1}{3}\)

Is this all the right inverses you're looking for?.

Note, if we sub \(\displaystyle a=\frac{-17}{18}, \;\ b=\frac{-4}{9}, \;\ c=\frac{1}{9}, \;\ d=\frac{1}{9}, \;\ e=\frac{13}{18}, \;\ f=\frac{2}{9}\) from [2], the system is satisfied.

Just a little input. Hope it helps a wee bit.