Question

obtain the inverse by using elementary row operation matrix is
9 5
2 7​

Answer 1

$\large\underline{\sf{Basic \: Concept - }}$

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.

$\large\underline{\bold{Solution-}}$

$\sf \: Let \: A \: = \: \begin{bmatrix} 9 & 5\\ 2 & 7\end{bmatrix}$

$\rm :\longmapsto\:A = IA$

$\rm :\longmapsto\: \: \begin{bmatrix} 9 & 5\\ 2 & 7\end{bmatrix} = \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}A$

$\bf :\longmapsto\:OP \: R_1 \to \: R_1 - 4R_2$

$\rm :\longmapsto\: \: \begin{bmatrix} 1 & - 23\\ 2 & 7\end{bmatrix} = \: \begin{bmatrix} 1 & - 4\\ 0 & 1\end{bmatrix}A$

$\bf :\longmapsto\:OP \: R_2 \to \: R_2 - 2R_1$

$\rm :\longmapsto\: \: \begin{bmatrix} 1 & - 23\\ 0 & 53\end{bmatrix} = \: \begin{bmatrix} 1 & - 4\\ - 2 & 9\end{bmatrix}A$

$\bf :\longmapsto\:OP \: R_2 \to \: \dfrac{1}{53} R_2$

$\rm :\longmapsto\: \: \begin{bmatrix} 1 & - 23\\ 0 & 1\end{bmatrix} = \: \begin{bmatrix} 1 & - 4\\ - \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}A$

$\bf :\longmapsto\:OP \: R_1 \to \: R_1 + 23R_2$

$\rm :\longmapsto\: \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} = \: \begin{bmatrix} \dfrac{7}{53} & - \dfrac{5}{53} \\ - \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}A$

Hence,

$\bf :\longmapsto\: {A}^{ - 1} = \: \begin{bmatrix} \dfrac{7}{53} & - \dfrac{5}{53} \\ - \dfrac{2}{53} & \dfrac{9}{53} \end{bmatrix}$