Using elementary row transformation ,find the inverse of the matrix
\begin{bmatrix} 3 & 0 &-1 \\ 2 & 3 &0 \\ 0&4 & 1 \end{bmatrix}.

 

 

 

 
 
 
 
 

Answers (1)

Let A= \begin{bmatrix} 3 & 0 &-1 \\ 2&3 &0 \\ 0& 4 &1 \end{bmatrix}
By row transformation , A=IA
A= \begin{bmatrix} 3 & 0 &-1 \\ 2&3 &0 \\ 0& 4 &1 \end{bmatrix}= \begin{bmatrix} 1 & 0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}A
By R_{1}\rightarrow R_{1}-R_{2}
\Rightarrow \begin{bmatrix} 1& -3 &-1 \\ 2&3 &0 \\ 0& 4 &1 \end{bmatrix}= \begin{bmatrix} 1 & -1 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}A
 By R_{2}\rightarrow R_{2}-2R_{1}
\Rightarrow \begin{bmatrix} 1& -3 &-1 \\ 0&9 &2 \\ 0& 4 &1 \end{bmatrix}= \begin{bmatrix} 1 & -1 &0 \\ -2&3 &0 \\ 0& 0 &1 \end{bmatrix}A
By R_{2}\rightarrow R_{2}-2R_{3}
\Rightarrow \begin{bmatrix} 1& -3 &-1 \\ 0&1 &0 \\ 0& 4 &1 \end{bmatrix}= \begin{bmatrix} 1 & -1 &0 \\ -2&3 &-2 \\ 0& 0 &1 \end{bmatrix}A
By  R_{3}\rightarrow R_{3}-4R_{2}
\Rightarrow \begin{bmatrix} 1& -3 &-1 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}= \begin{bmatrix} 1 & -1 &0 \\ -2&3 &-2 \\ 8& -12 &9 \end{bmatrix}A
By  R_{1}\rightarrow R_{1}+R_{3}
\Rightarrow \begin{bmatrix} 1& -3 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}= \begin{bmatrix} 9 & -13 &9 \\ -2&3 &-2 \\ 8& -12 &9 \end{bmatrix}A
By R_{1}\rightarrow R_{1}+3R_{2}
= \begin{bmatrix} 1& 0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}= \begin{bmatrix} 3 & -4 &3 \\ -2&3 &-2 \\ 8& -12 &9 \end{bmatrix}A
Hence A^{-1}= \begin{bmatrix} 3& -4 &3 \\ -2&3 &-2 \\ 8& -12 &9 \end{bmatrix}


 

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