[tex]\large\underline{\sf{Basic \: Concept – }}[/tex] Elementary Row Transformation Method :- The elementary transformation method to find the inverse of a matrix. To find the Inverse of a matrix, our goal is to convert the given matrix into an identity matrix. We can use three transformations:- 1) Multiplying a row by a constant 2) Adding a multiple of another row 3) Swapping two rows Basically, in elementary transformation of matrices we try to find out the inverse of a given matrix, using two simple properties : 1. A = A × I where I = Identity matrix 2. A × B =I it implies that B is inverse of A. [tex]\large\underline{\bold{Solution-}}[/tex] [tex] \sf \: Let \: A \: = \: \begin{bmatrix} 9 & 5\\ 2 & 7\end{bmatrix}[/tex] [tex]\rm :\longmapsto\:A = IA[/tex] [tex]\rm :\longmapsto\: \: \begin{bmatrix} 9 & 5\\ 2 & 7\end{bmatrix} = \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}A[/tex] [tex]\bf :\longmapsto\:OP \: R_1 \to \: R_1 – 4R_2[/tex] [tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & – 23\\ 2 & 7\end{bmatrix} = \: \begin{bmatrix} 1 & – 4\\ 0 & 1\end{bmatrix}A[/tex] [tex]\bf :\longmapsto\:OP \: R_2 \to \: R_2 – 2R_1[/tex] [tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & – 23\\ 0 & 53\end{bmatrix} = \: \begin{bmatrix} 1 & – 4\\ – 2 & 9\end{bmatrix}A[/tex] [tex] \bf :\longmapsto\:OP \: R_2 \to \: \dfrac{1}{53} R_2[/tex] [tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & – 23\\ 0 & 1\end{bmatrix} = \: \begin{bmatrix} 1 & – 4\\ – \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}A[/tex] [tex]\bf :\longmapsto\:OP \: R_1 \to \: R_1 + 23R_2[/tex] [tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} = \: \begin{bmatrix} \dfrac{7}{53} & – \dfrac{5}{53} \\ – \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}A[/tex] Hence, [tex]\bf :\longmapsto\: {A}^{ – 1} = \: \begin{bmatrix} \dfrac{7}{53} & – \dfrac{5}{53} \\ – \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}[/tex] Reply
[tex]\large\underline{\sf{Basic \: Concept – }}[/tex]
Elementary Row Transformation Method :-
We can use three transformations:-
Basically, in elementary transformation of matrices we try to find out the inverse of a given matrix, using two simple properties :
where I = Identity matrix
[tex]\large\underline{\bold{Solution-}}[/tex]
[tex] \sf \: Let \: A \: = \: \begin{bmatrix} 9 & 5\\ 2 & 7\end{bmatrix}[/tex]
[tex]\rm :\longmapsto\:A = IA[/tex]
[tex]\rm :\longmapsto\: \: \begin{bmatrix} 9 & 5\\ 2 & 7\end{bmatrix} = \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}A[/tex]
[tex]\bf :\longmapsto\:OP \: R_1 \to \: R_1 – 4R_2[/tex]
[tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & – 23\\ 2 & 7\end{bmatrix} = \: \begin{bmatrix} 1 & – 4\\ 0 & 1\end{bmatrix}A[/tex]
[tex]\bf :\longmapsto\:OP \: R_2 \to \: R_2 – 2R_1[/tex]
[tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & – 23\\ 0 & 53\end{bmatrix} = \: \begin{bmatrix} 1 & – 4\\ – 2 & 9\end{bmatrix}A[/tex]
[tex] \bf :\longmapsto\:OP \: R_2 \to \: \dfrac{1}{53} R_2[/tex]
[tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & – 23\\ 0 & 1\end{bmatrix} = \: \begin{bmatrix} 1 & – 4\\ – \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}A[/tex]
[tex]\bf :\longmapsto\:OP \: R_1 \to \: R_1 + 23R_2[/tex]
[tex]\rm :\longmapsto\: \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} = \: \begin{bmatrix} \dfrac{7}{53} & – \dfrac{5}{53} \\ – \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}A[/tex]
Hence,
[tex]\bf :\longmapsto\: {A}^{ – 1} = \: \begin{bmatrix} \dfrac{7}{53} & – \dfrac{5}{53} \\ – \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}[/tex]