Find the inverse of the following matrix, using elemebtary transformations:
A= \begin{bmatrix} 2 & 3 &1 \\ 2& 4 & 1\\ 3& 7 & 2 \end{bmatrix}

 

 

 

 
 
 
 
 

Answers (1)

A= \begin{bmatrix} 2 &3 &1 \\ 2& 4 &1 \\ 3& 7 &2 \end{bmatrix}  find A^{-1}  using elementary trasformation
A= IA\begin{bmatrix} 2 & 3 &1 \\ 2 & 4 &1 \\ 3& 7 &2 \end{bmatrix}= \begin{bmatrix} 1& 0 &0 \\ 0 & 1 &0 \\ 0& 0 &1 \end{bmatrix}A
          I= A^{-1}A
          R_{1}\rightarrow R_{1}+R_{3}
\begin{bmatrix} 1 & 4 &1 \\ 2 & 4 &1 \\ 3& 7 &2 \end{bmatrix}= \begin{bmatrix} -1 & 0 &1 \\ 0 & 1 &0 \\ 0& 0 &1 \end{bmatrix}A
R_{2}\rightarrow R_{2}-2R_{1}\: \: \S \: \: R_{3}\rightarrow R_{3}-3R_{1}
\begin{bmatrix} 1 & 4 &1 \\ 0 & -4 &-1 \\ 0& -5 &-1 \end{bmatrix}= \begin{bmatrix} -1 & 0 &1 \\ 2 & 1 &-2 \\ 3& 0 &-2 \end{bmatrix}A
R_{2}\rightarrow R_{2}-R_{5}
\begin{bmatrix} 1 & 4 &1 \\ 0 & 1 &0 \\ 0& 5 &-1 \end{bmatrix}= \begin{bmatrix} -1 & 0 &1 \\ -1 & 1 &0 \\ 3& 0 &-2 \end{bmatrix}A
R_{1}\rightarrow R_{1}-4R_{2}     &  R_{3}\rightarrow R_{3}+5R_{2}
\begin{bmatrix} 1 & 0 &1 \\ 0 & 1 &0 \\ 0& 0 &-1 \end{bmatrix}= \begin{bmatrix} 3 & -4 &1 \\ -1 & 1 &0 \\ -2& 5 &-2 \end{bmatrix}A
R_{3}\rightarrow -R_{3}
\begin{bmatrix} 1 & 0 &1 \\ 0 & 1 &0 \\ 0& 0 &1 \end{bmatrix}= \begin{bmatrix} 1 & 1 &-1 \\ -1 & 1 &0 \\ 2& -5 &2 \end{bmatrix}A
A^{-1}= \begin{bmatrix} 1 & 1 &-1 \\ -1 & 1 &0 \\ 2& 5 &2 \end{bmatrix}

      

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