Linear Functions and Matrices

Definition of Matrices:

A matrix is a group of numbers arranged into a fixed quantity of rows and columns. Usually a matrix is a two dimensional range of numbers or terms arranged in a set of rows and columns. m × n matrices A have m rows and n columns and it is written,

A = `[[a_(11),a_(12),a_(1n)],[a_(21),a_(22),a_(2n)],[a_(m1),a_(m2),a_(mn)]]`

Definition of Linear Functions:

A polynomial functions of single degree is defined as a linear functions. It relates a dependent variable with an independent variable in a simple way. Mathematical equation in which there is no independent-variable is raised to a power greater than one. A simple linear function with one independent variable traces a straight line when plotted on a graph. It is also called as linear equation.
Solving Linear Functions Using Matrices:

Steps to solve 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. So,

Hence, X = A−1C.
Example for Solving Linear Functions Using Matrices:

Example: 5x – y + 11 = 0 and x – y – 5 = 0, solve the linear equations for x and y using matrices.

Solution:           5x – y + 11 = 0     => 5x – y = – 11,

x – y – 5 = 0         => x – y = 5.

Let, A =` [[5,-1],[1,-1]]` , X = `[[x],[y]]` , and C = `[[-11],[5]]`

To find the value of A−1:

A = ` [[5,-1],[1,-1]]`

Swap the leading diagonal values,

=> ` [[-1,-1],[1,5]]`

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

=>` [[-1,1],[-1,5]]`

Now, find the value of |A|.

|A| = (–1 × 5) – ((–1) × 1)

=> |A| = –5 + 1

=> |A| = –4

Hence, A−1 = `1/(|A|) xx [[-1,1],[-1,5]]`

=> A−1 = `1/(-4) xx [[-1,1],[-1,5]]`

=> A−1 =` [[(-1)/(-4),1/(-4)],[(-1)/(-4),5/(-4)]]`

=> A−1 =` [[1/4,(-1)/4],[1/4,(-5)/4]]`

Therefore the solution is,

X = A–1C,

=> X =` [[1/4,(-1)/4],[1/4,(-5)/4]] xx [[-11],[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.