Question

Ques: 3x+y+2z=3 , 2x-3y-z= -3 , x+2y+z=4.
Solve the equation by matrix inversion method..

can u explain how -5 had come where i marked in the attached pic

Answer 1

Given : 3x+y+2z=3 , 2x-3y-z= -3 , x+2y+z=4.

To Find :  Solve the equation by matrix inversion method..

Solution:

$A=\left[\begin{array}{ccc}3&1&2\\2&-3&-1\\1&2&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$$=\left[\begin{array}{ccc}3\\-3\\4\end{array}\right]$

| A | = 8

A⁻¹ = (1/| A|  ) AdjA

$Adj A=\left[\begin{array}{ccc}C_{11}&C_{21}&C_{31}\\C_{12}&C_{22}&C_{32}\\C_{13}&C_{23}&C_{33}\end{array}\right]$

C₁₁ = (-1)¹⁺¹ (-3*1 - (2)(-1)) = - 1

C₁₂ = (-1)¹⁺² (2*1 - (1)(-1)) = - 3

C₁₃ = (-1)¹⁺³ (2*(2) - (1)(-3)) = 7

and so on..

C₂₃= (-1)²⁺³ ( 3 *2  - 1 * 1)

=> C₂₃= (-1)⁵ (6  -1)  

=>  C₂₃= (-1) (5)  

=>  C₂₃= -5

$A^{-1}=\dfrac{1}{8}\left[\begin{array}{ccc}-1&3&5\\-3&1&7\\7&-5&-11\end{array}\right]$

$\dfrac{1}{8}\left[\begin{array}{ccc}-1&3&5\\-3&1&7\\7&-5&-11\end{array}\right]\left[\begin{array}{ccc}3\\-3\\4\end{array}\right] =\left[\begin{array}{ccc}1\\2\\-1\end{array}\right]$

X = 1

Y = 2

Z =- 1

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