Showing posts with label Inverse. Show all posts
Showing posts with label Inverse. Show all posts

## HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD

How to find solutions of system of linear equations of three variables with  the help of Matrix Method. Let us solve three linear equations of three variables given below by Matrix Method. These are not simple linear equations to solve. But these equations can be made simple by following substitution .

After substitution , these equations are reduced to following simple linear equations
2x + 3y + 4z = 3
3x - 4y + 5z = 5
x + 2y - 3z = 6
To solve these equations by Matrix Method , Let us transform theses set of linear equations to Matrix form

AX = B,   and  =A-1 B---------------- (4)

where A is matrix comprises the coefficients  x , y and z respectively as shown in the matrix given below , And B is the matrix consist of constant terms on right hand side of these equations and X is the matrix of variables in the linear equations.

1st of all we have to find A- and then product of the matrices A-1 and  as mentioned in equation (4). But to find inverse of matrix A, its determinant value must be non zero   ( Note it) .
Let us find determinant value of Matrix A as follows :-

|A| = 2[(-4)(-3)–(5)(2)]-3[(3)(-3)-(5)(1)]+4[(3)(2)-(-4)(1)]

|A| = 2[12-10]-3[-9-5]+4[6+4]
|A| = 2 -3[-14]+4
|A| = 4 +42+40
|A| = 86
Since the value of Determinant of Matrix a is non zero.
⇒    A-1  exists .
To find inverse ,we have to find ad joint of Matrix A . And to find Ad joint of any Matrix , we have to find Co factor Matrix .

## What is Co Factor Matrix

Co factor Matrix can be obtained by the co factors of all the elements of Matrix written in same place  where the element was written originally.
Formula for Co factor of any element
Cij = (-1)i+jMij

### Co factor of A11

Co factor of A11   element can be calculated by eliminating 1st row and 1st column and solving the remaining determinant.
(-1)1+1M11
= 12 - 10 = 2

### Co factor of A12

Co factor of A12   element can be calculated   by eliminating 1st row and 2nd column and solving the remaining determinant.
(-1)1+2M12

= -1(-9 - 5 )= 14

### Co factor of A13

Co factor of A13   element  can be  calculated   by eliminating 1st row and 3rd column and solving the remaining determinant.
(-1)1+3M13
= 6 + 4 = 10

### Co factor of A21

Co factor of A21   element can be calculated   by eliminating 2nd row and 1st column and solving the remaining determinant.
(-1)2+1M11

= -1(- 9 - 8) = 17

### Co factor of A22

Co factor of A22   element  can be calculated   by eliminating 2nd row and 2nd column and solving the remaining determinant.
(-1)2+2M22
= - 6 - 4 = -10

### Co factor of A23

Co factor of A23   element can be calculated   by eliminating 2nd row and 3rd column and solving the remaining determinant.

(-1)2+3M23
= -1(4 - 3) = -1

### Co factor of A31

Co factor of A31   element can be calculated   by eliminating 3rd row and 1st column and solving the remaining determinant.

(-1)3+1M31
= 15 + 16 = 31

### Co factor of A32

Co factor of A32   element  can  be calculated   by eliminating 3rd row and 2nd column and solving the remaining determinant.

(-1)3+2M32
= -1(10 - 12) = 2

### Co factor of A33

Co factor of A33   element  can  be calculated   by eliminating 3rd row and 3rd column and solving the remaining determinant.

(-1)3+3M33
= - 8 - 9 = -17

## Co Factor Matrix of A

Now write all the co factors calculated in Matrix Form as follows

## Ad joint Matrix of A

To find the Ad Joint of  Co factor Matrix, transform 1st row to 1st column, 2nd row to 2nd column  and 3rd rows to 3rd column as follows

## Formula to Find   A-1

To Find the value of  A-1 ,Putting the value of Adj A in the formula given below.

## Find values of x ,y and z

Multiplying both the matrices which are on the right side.
Simplifying the matrix so obtained
Dividing each element by 86 to get matrix of 3×1 (i. e. Perform scalar multiplication of matrix ).

Using the equality of two matrices .We can write the values of x, y  and z respectively like this

x =277/86  , y = 4/86  and z = -77/86

Now the values of P, Q and R can be calculated by putting the values of x , y and z in equations (1) , (2 ) and (3) respectively.

P = 86/277  ,Q = 86/4   and R = -86/77

Don't Forget to Watch this Video of same Method

## Verification of solution

Putting values of P , Q and R in given  equations , these values must satisfies  three given equations
From 1st equation

Similarly other two equations can be verified

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## Conclusion

I have discussed the method of solving the System of linear Equations with the help of Matrix Method , method to find inverse of a matrix ,How to find inverse of 3×3 Matrix,how to solve determinant,how to solve system of equation by matrix,matrix method of solving system of equations of three variables,If you liked the post Don't  forget to share it with your friends , And in case of any improvement please make use of Comment Box .

## Appeal

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## 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 R1  R2  and  R3       respectively

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
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

## 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 - R2  and  R3 ➡️ R1 - R3

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 )]
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

Thanks for investing your precious time to read this post containing  Solving Determinants using elementary transformations,Short trick of elementary transformation  , elementary row transformation questions. If you liked it then share it with your near and dear ones to benefit them. we shall meet in next post with another beneficial article till then bye ,take care.......

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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 .

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

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

### 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

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

## Method for 3×3  Matrix Ist of all  Write the given Matrix in five columns by adding the 4th column as repetition of 1st column and 5th column as repetition of 2nd 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                        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                      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)

⎾ 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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