Solve the equation by inversion method :2x + 6y = 8, x + 3y = 5

1.	Solve the equation by inversion method :2x + 6y = 8, x + 3y = 5
Answer» Given, 2x + 6y = 8, x + 3y = 5 The given equations can be written in the matrix form as : \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \) \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix} 8 \\[0.3em] 5\\[0.3em] \end{bmatrix} \) This is of the form AX = B, where A = \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \), X = \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) and B = \( \begin{bmatrix} 8 \\[0.3em] 5\\[0.3em] \end{bmatrix} \) Let us find A^-1. \|A\| = \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \)= 6 – 6 = 0 ∴ A^-1 does not exist. Hence, x and y do not exist.

Answer»

Given,

2x + 6y = 8,

x + 3y = 5

The given equations can be written in the matrix form as :

\( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \) \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix} 8 \\[0.3em] 5\\[0.3em] \end{bmatrix} \)

This is of the form AX = B, where

A = \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \), X = \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) and B = \( \begin{bmatrix} 8 \\[0.3em] 5\\[0.3em] \end{bmatrix} \)

Let us find A^-1.

|A| = \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \)= 6 – 6 = 0

∴ A^-1 does not exist.

Hence,

x and y do not exist.

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