Finding the inverse of a matrix

[thelazyman]

Hi, I found the solved matrix from the original matrix


4 0 5
0 1 -6
3 0 4


and I managed to solve to this part


[1 0 5/4 \ 1/4 0 0]
[0 1 -6 \ 0 1 0]
[0 0 1 \ - 3 0 4]



The answer they got is

[1 0 0 \ 4 0 -5]
[0 1 0 \ -18 1 24]
[0 0 1 \ -3 0 4]


I tried solving this and I do not know how they managed to get that for the inverse of the matrix.

Thank you

 

[soroban]

Hello, thelazyman!

Find the inverse of: \(\displaystyle \:A \;=\;\begin{pmatrix}4&0&5 \\ \quad\\ 0&\:1\:&-6 \\ \quad\\ 3&0&4\end{pmatrix}\)

Advice: Try to avoid introducing fractions (if possible).

We have: \(\displaystyle \:\begin{vmatrix}4 & 0 & 5 & | & 1 & 0 & 0 \\
0 & 1 & 6 & | & 0 & 1 & 0 \\
3 & 0 & 4 & | & 0 & 0 & 1\end{vmatrix}\)


\(\displaystyle \begin{array}{ccccccccccc}R1-R3 \\ \quad \\ \quad \\ \quad \\ \quad \\ \quad \\ \quad \\ \quad \\ \quad \\ \quad \\ .\end{array}\;
\begin{vmatrix}1 & 0 & 1 & | & 1 & 0 & -1 \\
0 & 1 & 6 & | & 0 & 1 & 0 \\
3 & 0 & 4 & | & 0 & 0 & 1\end{vmatrix}\)


\(\displaystyle \begin{array}{cccccccccc} . \\ \quad \\ \quad \\ \quad \\\quad \\ \quad \\ \quad \\ \quad \\ \quad \\ R3-3R_1\end{array}\;
\begin{vmatrix}1 & 0 & 1 & | & 1 & 0 & -1 \\
0 & 1 & 6 & | & 0 & 1 & 0 \\
0 & 0 & 1 & | & -3 & 0 & 4\end{vmatrix}\)


\(\displaystyle \begin{array}{ccccccccc}R1-R3 \\ \quad \\ \quad\\ R2-6R3 \\ \quad \\ \quad \\ \quad \\ \quad\end{array}\;
\begin{vmatrix}1 & 0 & 0 & | & 4 & 0 & -5\\
0 & 1 & 0 & | & -18 & 1 & -24 \\
0 & 0 & 1 & | & -3 & 0 & 4\end{vmatrix}\)


Therefore: \(\displaystyle \:A^{-1} \;=\;\begin{pmatrix}4 & 0 & -5 \\ \quad \\ -18 & \:1\: & -24 \\ \quad \\ -3 & 0 & 4\end{pmatrix}\)