Solving Linear Equations Using Matrices

A matrix is a cluster of figures assembled into a predetermined magnitude of rows and columns. Regularly a matrix is a 2D array of figures or conditions arranged in a set of rows and columns. If matrices have m rows and n columns then it can be written as,

A = [[a_11,a_12,a_(1n)],[a_21,a_22,a_(2n)],[a_(m1),a_(m2),a_(mn)]]

Definition of Linear Equations:

The polynomial equations are said to be a linear equation, when the equation has the degree of one. A simple linear equations with one independent variable (y = a + bx) traces a straight line when plotted on a graph. It is also said to be linear function. In this article we shall discuss about solving linear equations using matrices.
Solving Linear Equations Using Matrices:

Steps for solving the system of linear equations using matrices,

Step 1: Let us consider the equation,

a1x + b1y = c1,

a2x + b2y = c2,

Step 2: If, A =[[a_1,b_1],[a_2,b_2]], X =[[x],[y]], and C =[[c_1],[c_2]].

then, AX = C.

Step 3: Now multiply both sides by, A−1

=> A−1AX = A−1C

Step 4: We know that, A−1A = I, where I is identity matrices,

=> IX = A−1C.

Step 5: Now, Multiplication of any matrices with identity matrices gives the same matrices.

Hence, X = A−1C.
Examples for Solving Linear Equations Using Matrices:

Example: Find the values of x and y by solving the linear equations, 3x – 2y + 5 = 0 and 6x + y + 5 = 0, using matrices.

Solution:3x – 2y + 5 = 0     => 3x – 2y = –5,

6x + y + 5 = 0       => 6x + y = –5.

Let, A =  [[3,-2],[6,1]], X =[[x],[y]] , and C =[[-5],[-5]]

To find the value of A−1:

A = [[3,-2],[6,1]]

Exchange the leading diagonal values,

=> [[1,-2],[6,3]]

Change the signs of the other two elements in the matrices,

=>  [[1,2],[-6,3]]

Now, find the determinant value of A, i.e. |A|.

|A| = (1 × 3) – (6 × (–2))

=> |A| = 3 + 12

=> |A| = 15

Hence, A−1 = 1/(|A|) xx [[1,2],[-6,3]]

=> A−1 = 1/(15) xx [[1,2],[-6,3]]

=> A−1 = [[1/15,2/15],[-6/15,3/15]]

=> A−1 = [[1/(15),2/(15)],[-2/5,1/5]]

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Therefore the solution is,

X = A–1C,

=> X = [[1/(15),2/(15)],[-2/5,1/5]]xx[[-5],[-5]]

=> X = [[(-5)/(15) -(10)/(15)],[(10)/5 - 5/5]]

=> X = [[(-15)/(15)],[5/5]]

=> X = [[-1],[1]]=[[x],[y]]

Hence the solutions are, x = – 1, y = 1.