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

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

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 R_1  , R_2  , and  R₃       respectively

If we add R_1    to   R_2  and   R₁   to  R_{3   then we get zero in 1st column, so}  Operating  R_{1  →} R₁   + R₂   and    R_{3  →} R₁   + R₃  

_{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

Proof:- Put L H S of determinant to Δ

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

Taking common xyz from C₃

Operating  R₂ ➡️ R₁ - R_2  and  R₃ ➡️ R₁ - R3

Expanding along  C₁ 

Δ = (x² - y² )( x³ - z³ ) - (x³ - y³ )(x² - z² )

Δ = (x- y )(x+ y)(x- z )( x² + z²  + xz ) - (x- y )( x² + y²  + xy ) (x - z )(x +z )

Δ = (x- y )(x- z )[(x+ y)( x² + z²  + xz ) -( x² + y²  + xy ) (x +z )]

Cancelling the same colour terms in the previous line ,then we have 

 Δ = (x- y )(x- z )[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 = (x- y )(x- z )[(yz² - y²z) +( xz²  - y²x)]

Δ = (x- y )(x- z )[yz(z - y) + x(z²  - y²)]

Δ = (x- y )(x- z )[yz(z - y) + x(z - y)(z+ y)]

Δ = (x- y )(x- z )(z - y)[yz + x(z+ y)]

Δ = (x- y )(x- z )(z - y)[yz + xz+ xy]

Taking -1 common from (x- z )(z - y) in previous line ,

Δ = (x- y )(y- z )(z - x)[yz + xz+ xy]

Hence the   proof

Also Read   HOW TO FIND THE SLOPE OF LINE  AX + BY = C ,  SLOPE OF A  LINE

Final Words

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HOW TO FIND THE INVERSE OF 2×2 AND 3×3 MATRIX USING SHORTCUT METHOD

Hello and Welcome to this post ,Today we are going to discuss the shortest and easiest methods of finding the Inverse of 2×2 matrix and 3×3 Matrix. Usually when we have to find the Inverse  of any  Matrix  then we follow the following steps .

1 Check whether the determinant value of the given Matrix is Non Zero.

 2  Find out   the co-factors of all the elements of the Matrix.

 3 Put these co-factors in co-factor Matrix.

 4 Find the Ad joint of this matrix by taking the  Transpose of a Matrix  of the co-factor matrix.

  5 Now  Multiply  Ad Joint of Matrix   with the reciprocal of               Determinant value of  the given Matrix.

This Method is very confusing, Long  and time Consuming.  So Let us have a New,  Easy and Shortcut Method .

Method For 2×2 Matrix

If we have to find the Inverse of  2×2 Matrix then Follows these steps.

 1 Interchange the position of the elements which are  a11  and a22 .

2  Change the Magnitude of the elements  which are in position a12 and  a21   .

3  Divide  every elements of the given Matrix with its Determinant value.

Example

 To find the Inverse of this matrix just interchange the position of elements a₁₁ and a₂₂  i.e   Interchange the positions of elements  5  and -3 and in second step change the magnitude of the elements which are  in positions a12 and  a21   i.e. change the sign of 9 and 4.

Now divide each elements with determinants value of the matrix which is  (5)(-3) - (9)(4) = -15 -36 = -51

So The Inverse of the given Matrix A  will  be 

Then after interchanging the positions of 8 and 2 change the magnitude of  7 and -6 and divide every elements with its determinant value (8)( 2) - (7)*(-6) = 16+42 = 58

The Inverse of  B is

 After interchanging the position of -3 and -6 and changing the magnitude of  -4 and -5 and at last dividing every elements with its determinant value (-3)×(-6) - (-4)×(-5) = 18 - 20 =  -2

The Inverse of  C is   

This video Explains all about Inverse of 2×2 and 3×3 Matrix

READ  MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS

Method for 3×3  Matrix        

Ist of all  Write the given Matrix in five columns by adding the 4^th column as repetition of 1^st column and 5^th column as repetition of 2^nd column, then

C₁    C2     C3      C4     C5
5      -1       4       5      -1
2       3       5       2        3
5      -2      6        5      -2

Now Expanding this Matrix to 5×5 Matrix by adding 4th Row as repetition of 1st Rows and adding 5 Row as repetition of 2nd column as what we received in last step.

R₁        5             -1             4            5          -1

R₂        2              3              5           2            3

R₃        5             -2             6            5          -2

R₄        5             -1             4            5          -1

R5        2              3              5           2            3

 Now to find the Inverse  of the given Matrix ,we have to find the cofactor of every elements

1 Find the co-factor of 1st element of Row 1 i. e. 5, determinant value of the Matrix (RED below ) obtained by eliminating the 1st Row and 1st Column which will be (3×6)-{(5)×(-2)} = 28,write these co-factor value in 1st column of 1st Row. (we are evaluating co-factors row wise and writing Column wise)

R₁      5      -1      4        5          -1
R₂    2       3      5        2           3
R₃    5      -2      6       5          -2
R₄    5      -1      4       5          -1
R5     2       3      5       2           3

2 Now Find the co-factor of 2nd element of 1st Row i. e. -1,which is equal to determinant value of the Matrix  (RED below) obtained by eliminating the 1st Row and 2nd Column which will be 5*5-(2)*(6) =13,write this co-factor value in  2nd Row of 1st column .(we are evaluating co-factors row wise and writing Column wise) 

R₁        5             -1             4            5          -1

R₂        2              3              5           2            3

R₃        5             -2             6            5          -2

R₄        5             -1             4            5          -1

R5        2              3              5           2            3 

3 Now Find the co-factor of 3rd element of 1st Row  i.e. 4, which is equal to determinant value of the Matrix  (RED below ) obtained by eliminating the 1st Row and 3rd Column which will be 2*(-2)-(3)*(5) = -19,write this co-factor value in  3rd Row of 1st column.(we are evaluating co factors row wise and writing Column wise)

R₁           5            -1            4             5         -1

R₂           2             3             5            2           3

R₃           5            -2             6            5          -2

R₄           5            -1             4            5          -1

R5           2             3             5            2           3

4  Now Find the co-factor of 1st element of 2nd Row  i. e. 2, which is equal to determinant value of the Matrix  (RED below ) obtained by eliminating the 2nd  Row and 1st Column which will be -2*(4)-(6)*(-1) = -2,write this co-factor value in  2nd Column of 1st Row .(we are evaluating co factors row wise and writing Column wise) .

R₁         5         -1           4             5           -1

R₂         2          3            5             2            3

R₃         5         -2            6            5           -2

R₄         5         -1            4            5           -1

R5         2          3            5            2             3

5 Now Find the co-factor of 2nd element of 2nd Row i.e 3, which is equal to determinant value of the matrix (RED below ) obtained by eliminating the 2nd Row and 2nd column which will be 6*(5)-(5)*(4) = 10,write this co-factor value in 2nd Row of 2nd column

  R₁        5           -1          4          5          -1

  R₂        2            3           5         2           3

  R₃        5           -2          6          5          -2

  R₄        5           -1          4          5          -1

  R5        2            3          5          2           3

6 Find the co-factor of 3rd element of 2nd Row i. e.5, which is equal to determinant value of the Matrix (RED ) obtained by eliminating the 2nd Row and 2nd Column which will be 5× (-1)-(-2) × (5) = 5,write this co-factor value this 2nd Column of 3rd Row, write this co-factor value in 2nd Column of 1st Row . (we are evaluating co factors row wise and writing Column wise ) .

 R₁         5          -1           4           5          -1

 R₂         2           3           5           2            3

 R₃         5          -2           6           5          -2

 R₄         5          -1           4           5          -1

 R5         2           3           5            2           3

Similarly for 1st , 2nd, 3rd element the co-factor values will be as follows

      For  A₃1 i.e  5

R₁       5            -1             4            5          -1

R₂       2             3              5           2           3

R₃       5            -2             6            5          -2

R₄       5            -1             4            5          -1

R5       2             3              5           2           3

For A₃₂ i.e   -2
R₁          5         -1            4            5          -1 
R2          2          3            5            2           3

R₃          5         -2            6            5          -2

R₄          5          -1           4            5          -1

R5          2           3           5            2           3

  for  A₃₃  i.e. 6

R₁          5           -1           4           5          -1

R₂          2            3           5           2           3

R₃          5           -2           6           5         -2

R₄          5           -1           4           5          -1

R5          2            3           5           2           3 

so we have  -17 ,-17 and 17 as co-factors of 3rd Row, write these co factors in 3rd column .

(we are evaluating co factors row wise and writing Column wise)

Ad joint   A   =   

                                        
⎾ 28          -2          -17 ⏋
⎹⎸ 13          10         -17 ⎹
⎿ -19         5            17  ⏌

Now divide with the determinant value of given 3×3 Matrix , which will be 5(28)-1(-13) + 4(-19) = 140 + 13 -7 6 = 77.

Now divide each element of Ad joint Matrix obtained in previous step with determinant value 77,

Then   A⁻¹  =   

Conclusion

This  post was regarding short cut methods of finding Inverse of  2×2 and 3×3 Matrices , If you liked this post ,Please  share your precious views on this topic and share this post with your friends to benefit them. we shall Meet in the next post ,till then BYE . 

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