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1/16x+1/15y=9/20 , 1/20x-1/27y=4/45 Related: Equations and Matrices (...
Equations and Matrices (Part - 1)


Introduction

In this topic, we will be discussing how to solve systems of linear equations using matrices. We will be using the method of elimination to reduce the system of equations to an upper triangular matrix, which can then be solved using back-substitution.


System of Equations

The given system of equations are:

1/16x + 1/15y = 9/20

1/20x - 1/27y = 4/45


Matrix Form

We can write the given system of equations in matrix form as:

AX = B

where,

A = [1/16 1/15; 1/20 -1/27]

X = [x;y]

B = [9/20; 4/45]


Augmented Matrix

We can write the augmented matrix for the given system of equations as:

[1/16 1/15 | 9/20; 1/20 -1/27 | 4/45]


Reduced Row Echelon Form

Using the method of elimination, we can reduce the augmented matrix to its reduced row echelon form:

[1 0 | 4/3; 0 1 | -2/3]


Solution

From the reduced row echelon form, we can see that:

x = 4/3

y = -2/3


Conclusion

Thus, the solution to the given system of equations is x = 4/3 and y = -2/3.
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1/16x+1/15y=9/20 , 1/20x-1/27y=4/45 Related: Equations and Matrices (Part - 1)?
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