Find the inverse of the following matrix, using elementary transformations:

A=\begin{bmatrix} 2 &0 &-1 \\ 5& 1 & 0\\ 0& 1 & 3 \end{bmatrix}

 

 

 

 

 
 
 
 
 

Answers (1)

Given, A=\begin{bmatrix} 2 &0 &-1 \\ 5& 1 & 0\\ 0& 1 & 3 \end{bmatrix}

we know that 

AA^{-1}=I

\therefore A=IA

\begin{bmatrix} 2 &0 &-1 \\ 5& 1 & 0\\ 0& 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1 & 0\\ 0& 0 & 1 \end{bmatrix}A

On applying R_3\rightarrow R_3+3R_1

\begin{bmatrix} 2 &0 &-1 \\ 5& 1 & 0\\ 6& 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1 & 0\\ 3& 0 & 1 \end{bmatrix}A

Applying R_3\rightarrow R_3-R_2

\begin{bmatrix} 2 &0 &-1 \\ 5& 1 & 0\\ 1& 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1 & 0\\ 3& -1 & 1 \end{bmatrix}A

Interchanging R_1\leftrightarrow R_3

\begin{bmatrix} 1&0 &0 \\ 5& 1 & 0\\ 2& 0 & -1 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ 0& 1 & 0\\ 1& 0 & 0 \end{bmatrix}A

Applying R_2\rightarrow R_2-5R_1\: \: and\: \: R_3\rightarrow R_3-2R_1

\begin{bmatrix} 1&0 &0 \\ 0& 1 & 0\\ 0& 0 & -1 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ -15& 6 & -5\\ -5& 2 & -2 \end{bmatrix}A

Applying R_3\rightarrow (-1)R_3\begin{bmatrix} 1&0 &0 \\ 0& 1 & 0\\ 0& 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ -15& 6 & -5\\ 5& -2 & 2 \end{bmatrix}A

Hence the required inverse of the matrix is

A^{-1}=\begin{bmatrix} 3 &-1 &1 \\ -15& 6 & -5\\ 5& -2 & 2 \end{bmatrix}

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