Obtain the inverse of the following matrix using elementary operation:
A= \begin{bmatrix} -1 &1 &2 \\ 1&2 &3 \\ 3&1 & 1 \end{bmatrix}

 

 

 

 
 
 
 
 

Answers (1)

given\, A= \begin{bmatrix} -1 & 1 &2 \\ 1& 2 & 3\\ 3&1 & 1 \end{bmatrix}
A= IA Applying R_{1}\leftrightarrow R_{2}, we\, get
\begin{bmatrix} 1 & 2 &3 \\ -1& 1 & 2\\ 3&1 & 1 \end{bmatrix}= \begin{bmatrix} 0 & 1 &0 \\ 1& 0& 0\\ 0&0 & 1 \end{bmatrix}A
Applying R_{3}\rightarrow R_{3}-3R_{1}, weget
     Applying R_{2}\rightarrow R_{3}+R_{2}, weget
\begin{bmatrix} 1 & 2 &3 \\ 0& 3 & 5\\ 0&-5& -8 \end{bmatrix}= \begin{bmatrix} 0 & 1 &0 \\ 1& 1& 0\\ 0&-3 & 1 \end{bmatrix}A
Applying R_{2}\rightarrow \frac{1}{3}R_{2}
\begin{bmatrix} 1 & 2 &3 \\ 0& 1 & \frac{5}{3}\\ 0&-5 & -8 \end{bmatrix}= \begin{bmatrix} 0 & 1 &0 \\ \frac{1}{3}& \frac{1}{3}& 0\\ 0&-3 & 1 \end{bmatrix}A
Applying R_{1}\rightarrow R_{1}-2R_{2}
\begin{bmatrix} 1 & 0 &\frac{-1}{3} \\ 0& 1 & \frac{5}{3}\\ 0&-5 & -8 \end{bmatrix}= \begin{bmatrix} \frac{-2}{3} & \frac{1}{3} &0 \\ \frac{1}{3}& \frac{1}{3}& 0\\ 0&-3 & 1 \end{bmatrix}A
Applying R_{3}\rightarrow R_{3}+5R_{2}
\begin{bmatrix} 1 & 0 &\frac{-1}{3} \\ 0& 1 & \frac{5}{3}\\ 0&0 & \frac{1}{3} \end{bmatrix}= \begin{bmatrix} \frac{-2}{3} & \frac{1}{3} &0 \\ \frac{1}{3}& \frac{1}{3}& 0\\ \frac{5}{3}&\frac{-4}{3} & 1 \end{bmatrix}A
R_{1}\rightarrow R_{1}+R_{3}
\begin{bmatrix} 1 & 0 &0 \\ 0& 1 & \frac{5}{3}\\ 0&0 & \frac{1}{3} \end{bmatrix}= \begin{bmatrix} 1& -1 &1 \\ \frac{1}{3}& \frac{1}{3}& 0\\ \frac{5}{3}&\frac{-4}{3} & 1 \end{bmatrix}A
R_{2}\rightarrow R_{2}-5R_{3}
\begin{bmatrix} 1 & 0 &0 \\ 0& 1 & 0\\ 0&0 & \frac{1}{3} \end{bmatrix}= \begin{bmatrix} 1 & -1 &0 \\ -8& 2& -5\\ \frac{5}{3}&\frac{-4}{3}& 1 \end{bmatrix}A
R_{3}\rightarrow 3R_{3}

\begin{bmatrix} 1 & 0 &0 \\ 0& 1 & 0\\ 0&0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1 &0 \\ -8& 7& -5\\ 5&-4& 3 \end{bmatrix}A
I= A^{-1}A

\therefore A^{-1}= \begin{bmatrix} 1 & -1 &0 \\ -8& 7& -5\\ 5&-4 & 3\end{bmatrix}

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