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  4. Using elementary transformation, the inverse of the matrix [1 2 3 2 5 7 -2 -4 -5] is

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Q. Using elementary transformation, the inverse of the matrix $\begin{bmatrix}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{bmatrix}$ is

Matrices

Solution:

Let $A=\begin{bmatrix}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{bmatrix}$
Consider $A = IA$
$\therefore \begin{bmatrix}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A$
Applying $R _{2} \rightarrow R _{2}-2 R _{1},\, R _{3} \rightarrow R _{3}+2 R _{1}$, we get
$\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1\end{bmatrix}A$
Applying $R _{1} \rightarrow R _{1}-3 R _{3},\, R _{2} \rightarrow R _{2}- R _{3}$, we get
$\begin{bmatrix}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}= \begin{bmatrix}3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1\end{bmatrix}A$
Applying $R _{1} \rightarrow R _{1}-2 R _{2}$, we get
$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}= \begin{bmatrix}3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1\end{bmatrix} A$
$\therefore A^{-1}= \begin{bmatrix}3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1\end{bmatrix}$