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
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
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 = 4and 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
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 A12 = -5
co-factor of A13 = 1
Co-Factors of 2nd Row
co-factor of A21 = 2
co-factor of A22 = 0co-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
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.