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

INVERSE OF 2×2 AND 3×3 MATRIX

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.INVERSE OF 2×2 AND 3×3 MATRIX, shortcut to find inverse of 2 × 2 matrix,transpose 3x3,inverse of determinant,matrices and determinants, determinant calculator, crammer's rule.

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  =   

How to find Inverse of 2×2 Matrix using elementary Transformations.Short cut to inverse of 2×2 .

Conclusion 

This post was regarding short cut methods of finding Inverse of 2×2 and 3×3 Matrices , INVERSE OF 2×2 AND 3×3 MATRIX, shortcut to find inverse of 2 × 2 matrix, transpose 3x3,  inverse of determinant, matrices and determinants, determinant calculator, crammer's rule,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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HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES BY MATRIX METHOD

Today we are going to learn, how to solve system of linear equations of 2 and 3 variables using matrix method. So using matrix method to solve system of linear equation , we must know some topics such as co-factor of element, Transpose of matrix , Ad joint of a Matrix, Multiplication of two Matrices ,Determinant value of a Matrix ,Inverse of a matrix etc. Let us understand the process of finding the solution of system of linear equations with the help of some examples.

To Solve the system  of Linear Equations using 2×2 Matrix Method

x - 5y = 4

2x + 5y = −2

Writing this system of equation in Matrix form

AX = B 

where X = A-1 B------------------------(1)


And A-1 = (1/Det A ) ( Ad joint A)



Where

We need the inverse of   A , which we write as   A^-1 

To find the inverse 1st find out Co-factor Matrix of A

Ad joint  A  =   (     5    5​ -2   1   ​)

Co factor   A  =   (     5    −2​ 5   1   ​)

As we know the Ad joint Matrix of any matrix can be found by taking the transpose of the Co Factor matrix.

Now let us find the determinant of  A

∣A∣ = 5 − (-10) = 15 , which is non Zero,


Therefore  A^-1  Exists     ,So

Now putting the value of inverse of Matrix A in equation (1)

Now putting the elements of Matrix X , and 

By the equality of two Matrices ,their elements in respective positions are equal to each others,

Hence x= 2/3  and y = -2/3

How to check the correctness of a solution

put x=  2/3  and y = -2/3  in the given  equations  

x - 5y = 4 implies 2/3 - 5(-2/3) = 2/3+10/3 = 4

and 2x + 5y  = -2 implies 2*(2/3) + 5(-2/3) = 4/3-10/3 = -6/3= -2

therefore both equations are satisfied by these values of 'x' and 'y'

To Solve the system of Linear Equations using 3×3 Matrix Method

 x  - y + z   = 4
2x + y - 3z  = 0
 x + y + z    = 2

Converting this system of Linear equations into Matrix Form

AX    =  B 

X  =    A^-1  B ----------------------------------(1)
Where

Now we have to find the   A^-1 . , and it will exist if its determinants value |A| must be non Zero. Let us find |A|

    




 |A|= 1{1×1-(-3)×1}-(-1){2×1-(3)×1}+1(2×1-1×1)

|A|= 1 {1+3}+1 {2+3}+1 {2-1}

|A|= 4+5+1 = 10

Since |A| is non Zero , Therefore inverse of A exists .

We have to find the Co factors of all the elements of matrix A . Let us find out the co factors of all the elements of matrix row wise.

Co-Factors of 1st Row

     

            co-factor of A11  =   4
            co-factor of A12  =  -5
            co-factor of A13  =   1

Co-Factors of 2nd Row

            co-factor of A21  =   2 

            co-factor of A22  =   0
            co-factor of A23  =  -2

Co-Factors of 3rd Row

              co-factor of A31  =  2
              co-factor of A32  =  5
              co-factor of A33  =  3

Now co Factor matrix of A can be written as follows




Now to find the Ad joint of this Matrix, Take the transpose of this Matrix, 

As we know the Inverse of a Matrix  is the scalar  multiplication of Ad Joint Matrix of Matrix A and reciprocal of Determinant value of the Matrix A.

Putting the values of Inverse of A and Matrix B in equation (1), So after Multiplication of these two matrices ,we get

$\backslash displaystyle=\{\backslash left(\backslash begin\{matrix\}\{22\}\backslash \backslash -\{16\}\backslash \backslash -\{16\}\backslash end\{matrix\}\backslash right)\}$

Now putting the values of elements  of   Matrix X, and equating the elements in their respective positions .

Now using the property of equality of two Matrices , All the elements in their respective position are equal to each other , we get

 x = 2  , y = -1, z = 1  is the solution of the system of linear Equations.

Conclusion

In this post I  discussed the method of solving the System of linear Equations with the help of Matrix Method , If you liked the post please share it with your friends , And in case of any improvement please make use of Comment Box . To keep on supporting me follow my Blog . Thanks for your valuable time to read this post .We shall meet you in next post ,till then BYE.

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