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Using elementary row transformations, find the inverse of the matrix A=$$\begin{bmatrix}1&2&3 \\ 2&5&7 \\ -2&-4&-5 \end{bmatrix}$$.
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## QuestionMathsClass 12

Using elementary row transformations, find the inverse of the matrix A=$$\begin{bmatrix}1&2&3 \\ 2&5&7 \\ -2&-4&-5 \end{bmatrix}$$.

A=$$\begin{bmatrix}1&2&3 \\ 2&5&7 \\ -2&-4&-5 \end{bmatrix}$$
$$|A| = 1(-25+28)-2(-10+14)+3(-8+10)$$
$$=3-2(4)+3(2)$$
$$=9-8$$
$$=1≠0$$
$$A^{-1}$$exists.
$$A.A^{-1} = I$$
$$\begin{bmatrix}1&2&3 \\ 2&5&7 \\ -2&-4&-5 \end{bmatrix}A^{-1}=\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$$
$$\ce R_2\to \ce R_2-2\ce R_1$$;$$\ce R_3\to \ce R_3+2\ce R_1$$
$$\begin{bmatrix}1&2&3\\0&1&1\\0&0&1\end{bmatrix}A^{-1} = \begin{bmatrix}1&0&0\\-2&1&0\\2&0&1\end{bmatrix}$$
$$\ce R_1\to \ce R_1-2\ce R_2$$
$$\begin{bmatrix}1&0&1\\0&1&1\\0&0&1\end{bmatrix}A^{-1} = \begin{bmatrix}5&-2&0\\-2&1&0\\2&0&1\end{bmatrix}$$
$$\ce R_1\to \ce R_1-\ce R_3$$
$$\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}\ce A^{-1} = \begin{bmatrix}3&-2&-1 \\ -2&1&0 \\ 2&0&1 \end{bmatrix}$$
$$\ce R_2\to \ce R_2-\ce R_3$$
$$\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}\ce A^{-1} = \begin{bmatrix}3&-2&-1 \\ -4&1&-1 \\ 2&0&1 \end{bmatrix}$$
$$\ce I.\ce A^{-1} = \ce A^{-1} = \begin{bmatrix}3&-2&-1 \\ -4&1&-1 \\ 2&0&1 \end{bmatrix}$$