Mathematics of Linear Algebra

A finite set of equation with the variables x 1 , x 2 - - - - -x n is called a System of Linear Equation. a 1 x 1 + a 2 x 2 + - - - - - - + a n x n = b ----------------------- (1) Where a 1, a 2 - - - - - - a n and b are real constants. An equation of this form is called Linear Equation in the variables x 1 , x 2 - - - - - - x n . The variables in a linear equation are sometimes called Unknown. And there are two types of linear equation such as Homogeneous Linear Equation and Non-homogeneous Linear Equation. If b = 0 then (1) is called Homogeneous Linear Equation and if b is not equal 0 the (1) is called Non-homogeneous Linear Equation.

First Step: - Consider the following system of m linear equations (or set of m simultaneous linear equations) in n unknown x 1 , x 2 - - - - - - x n. a 11 x 1 + a 12 x 2 + - - - - - - + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + - - - - - - + a 2n x n = b 2 … …. …. ….. … … .. .. . .. . .. .. . .. .. .. .. . .. System (1) . … . .. .. . .. .. . .. .. . .. . .. . .. .. . .. . .. . .. .. a m1 x 1 + a m2 x 2 + - - - - - - + a mn x n = b m We reduce the System (1) to a simpler system as follows: Step 1: - Elimination of x 1 from the second, third……..mth equations. We may assume that the order (rule) of the equations and the order (rule) of the unknowns in each equation such that a 11 is not equal 0. The variables x 1 can then be eliminated from the second, third ………………..mth equations by subtracting. a 21 /a 11 times the first equation from the second equation a 31 /a 11 times the first equation from the third equation .. ...

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination 2x – y – 3z = 0 -x +2y – 3z = 0 x + y + 4z = 0 Solution: - The given system is 2x – y – 3z = 0 -x +2y – 3z = 0 System (1) x + y + 4z = 0 Let us represent the three linear equation of the system (1) by L1, L2, and L3 respectively. Reduce the system to echelon form by elementary operation. Apply L1 ß à L2 x + y + 4z = 0 -x +2y – 3z = 0 System (2) 2x – y – 3z = 0 Apply L2 à L2 +L1 and L3 à L3 – 2L1 Thus we obtain the equivalent system is x + y + 4z = 0 3y + z = 0 System (3) –3 y – 11z = 0 Now apply L3 à L3 + L2 Thus we obtain the equivalent system is ...

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination Solution: - The given system is 2I 1 – I 2 + 3I 3 + 4 I 4 = 9 I 1 - 2I 3 + 7I 4 = 11 System (1) 3I 1 – 3I 2 + I 3 + 5I 4 = 8 2I 1 + I 2 + 4I 3 + 4I 4 = 10 Let us represent the four linear equation of the system (1) by L 1 , L 2 , L 3 and L 4 respectively. Reduce the system to echelon form by elementary operation. Apply L 2 → L 1 – 2L 2 , L 3 → 3L 1 – 2L 3 and L 4 → L 4 – L 1 Thus we obtain the equivalent system is 2I 1 – I 2 + 3I 3 + 4I 4 = 9 – I 2 - 7I 3 - 10I 4 = 13 System (2) 3I 2 + 7 I 3 + 2I 4 = 11 2I 2 + I 3 + = 1 Apply L 3 → 3L 2 + L 3 and L 4 → 2L 2 + L 4 Thus we obtain the equivalent system is 2I 1 –...