HOW TO SOLVE DETERMINANTS USING ELEMENTARY TRANSFORMATIONS


How to Prove Determinants using elementary transformations 

In this post we shall discuss Short trick of elementary transformation,Solving Determinants using elementary transformations,define elementary transformation, elementary transformation class 12, elementary row transformation questions.


By this method we have  to reduce maximum elements of specific Rows or column to zero, so that we can solve it easily
How to solve determinants using elementary operations

To solve the determinants using elementary transformations , Let us suppose L H S = △
How to solve determinants using elementary operations



As we can see that 'a' is common in 1st Row , 'b' is common in 2nd Row and 'c' is common in 3rd row ,
Therefore Taking a ,b ,c common from R1  R2  and  R3       respectively


How to solve determinants using elementary operations

If we add R1    to   R2  and   R1   to  R3   then we get zero in 1st column, so  Operating  R1  → R1   + R2   and    R3  → R1   + R3  
How to solve determinants using elementary operations
As we have received maximum possible  zero in 1st column Therefore Expanding along C1  

△= (-a)×[(0)-(2c×2b)]
△ = abc{-a(-4bc)}

 4a2b2c2 

Hence the proof


Watch this video for Understanding Elementary transformations



PROBLEM

How to solve determinants using elementary operations

Proof:- Put L H S of determinant to Δ
How to solve determinants using elementary operations

Operating R1 ➡️xR1  , R2 ➡️ yR2 and R3➡️ zR3

How to solve determinants using elementary operations

Taking common xyz from C3

How to solve determinants using elementary operations
Operating  R2 ➡️ R1 - R2  and  R3 ➡️ R1 - R3

How to solve determinants using elementary operations

Expanding along  C1 



Δ = (xy2 )( xz3 ) - (xy3 )(xz2 )

Δ = (xy )(xy)(xz )( xz2  + xz ) - (xy )( xy2  + xy ) (x z )(x +z )

Δ = (xy )(xz )[(xy)( xz2  + xz ) -( xy2  + xy ) (x +z )]
HOW TO SOLVE DETERMINANTS USING ELEMENTARY TRANSFORMATIONS
Cancelling the same colour terms in the previous line ,then we have 
 Δ = (xy )(xz )[xz2   + yz2 - y2x - y2z  ]

Arranging  terms in Squared Bracket  in such a way that the term containing z2 must be at 1st and 3rd position and the term containing y2 must be at 2nd and 4th position .

Δ = (xy )(xz )[(yz2 - y2z) +( xz2  - y2x)]
Δ = (xy )(xz )[yz(z - y) + x(z2  - y2)]
Δ = (xy )(xz )[yz(z - y) + x(z - y)(zy)]
Δ = (xy )(xz )(z - y)[yz + x(zy)]
Δ = (xy )(xz )(z - y)[yz + xz+ xy]
Taking -1 common from (xz )(z - y) in previous line ,
Δ = (xy )(yz )(z - x)[yz + xz+ xy]

Hence the   proof

Final Words

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