Using elementary row transformations, find the inverse of the matrix 
\begin{bmatrix} 2 & -3 & 5\\ 3&2 &-4 \\ 1 &1 & -2 \end{bmatrix}

 

 

 

 
 
 
 
 

Answers (1)

Let A= \begin{bmatrix} 2 &-3 & 5\\ 3& 2 & -4\\ 1& 1 &-2 \end{bmatrix}
we know that A=IA\; so, \begin{bmatrix} 2 &-3 & 5\\ 3& 2 & -4\\ 1& 1 &-2 \end{bmatrix}= \begin{bmatrix} 1 &0 & 0\\ 0& 1 & 0\\ 0& 0 &1 \end{bmatrix}A
   by\: R_{1}\rightarrow R_{1}-R_{3}
\begin{bmatrix} 1 & -4&7 \\ 3& 2 & -4\\ 1&1 & -2 \end{bmatrix}= \begin{bmatrix} 1 & 0&-1 \\ 0& 1 & 0\\ 0&0 & 1 \end{bmatrix}A
By R_{2}\rightarrow R_{2}-3R_{1},\: R_{3}\rightarrow R_{3}-R_{1}
\begin{bmatrix} 1 & -4&7 \\ 0& 14 & -25\\ 0&5 & -9 \end{bmatrix}= \begin{bmatrix} 1 & 0&-1 \\ -3& 1 & 3\\ -1&0 & 2 \end{bmatrix}A
By\: R_{2}\rightarrow -\left ( R_{2}-3R_{3} \right )
\begin{bmatrix} 1 & -4&7 \\ 0& 1 & -2\\ 0&5 & -9 \end{bmatrix}= \begin{bmatrix} 1 & 0&-1 \\ 0& -1 & 3\\ -1&0 & 2 \end{bmatrix}A
By R_{1}\rightarrow R_{1}+4R_{2}\: ,R_{3}\rightarrow R_{3}-5R_{2}
\begin{bmatrix} 1 & 0&-1 \\ 0& 1 & -2\\ 0&0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -4&11 \\ 0& -1 & 3\\ -1&5 & -13 \end{bmatrix}A
R_{1}\rightarrow R_{1}+R_{3}\: ,R_{2}\rightarrow R_{2}+2R_{3}
\begin{bmatrix} 1 & 0&0 \\ 0& 1 & 0\\ 0&0 & 1 \end{bmatrix}= \begin{bmatrix} 0 & 1&-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}A
since I= A^{-1}A
\therefore A^{-1}= \begin{bmatrix} 0 & 1 &-2 \\ -2& 9 &-23 \\ -1& 5 &-13 \end{bmatrix}



 

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