Mathematics of Linear Algebra

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 –

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination 2x1 + 3x2 + 5x3 – x4 = 8 3x1 + 4x2 + 2x3 – 3x4 = 8 x1 + 2x2 + 8x3 – x4 = 8 7x1 + 9x2 + x3 – 8x4 = 8 Solution: - The given system is 2x1 + 3x2 + 5x3 – x4 = 8 3x1 + 4x2 + 2x3 – 3x4 = 8 System (1) x1 + 2x2 + 8x3 – x4 = 8 7x1 + 9x2 + x3 – 8x4 = 8 Let us represent the three 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 1 ↔L 3 Then we have the equivalent system x1 + 2x2 + 8x3 – x4 = 8 3x1 + 4x2 + 2x3 – 3x4 = 8 System (2) 2x1 + 3x2 + 5x3 – x4 = 8 7x1 + 9x2 + x3 – 8x4 = 8 Now Apply L 2 → L 2 - 3L 1 , L

For which value of a will be following system have no solution? Exactly one solution? Infinitely many solutions. x + 2y- 3z = 4 3x - y + 5z = 2 4x + y +(a 2 - 14)z = (a+2) Solution: - The given system is x + 2y- 3z = 4 3x - y + 5z = 2 System (1) 4x + y +(a 2 - 14)z = (a+2) Let us represent the three linear equation of the system (1) by L 1 , L 2 , and L 3 respectively. Reduce the system to echelon form by elementary operation. Apply L 2 → L 2 - 3L 1 and L 3 → L 3 – 4L 1 Thus we obtain the equivalent system is x + 2y- 3z = 4 - 7y + 5z = -10 System (2) - 7y + (a 2 - 2) z = (a -14) Now apply L 3 → L 3 - L 2 Thus we obtain the equivalent system is x + 2y- 3z = 4 - 7y + 5z = -10 System (3)