Assessment 1. 2. Solve the linear system of equations x — 3y — 6 5x+ 3'= 42 r. A) Unique solution: (6,4) B) Unique solution: Infinitely many solutions: (t, 8t + ) D) No solution Solve the linear...

1) x – 3y = -6 ---------eq1 (eq-equation)
5x + 3y = 42 --------- eq2
eq1 * 5 => 5x – 15y = -30
eq2 => 5x + 3y = 42
eq2 - eq1 * 5 = 18y = 72 => y=4
substitute y=4 in eq1 => x – (3*4) = -6 => x = 6
option A
2) 2x – 6y = 7 ---------eq1 (eq-equation)
5x + 2y = 10 --------- eq2
eq1 * 5 => 10x – 30y = 35
eq2 * 2 => 10x + 4y = 20
eq2*2 - eq1 * 5 = 34y = -15 => y = -15/34
substitute y = -15/34 in eq1 => 2x – (6* (-15/34)) = 7 => x = 37/17
option B
3) x + 16y = 9 ---------eq1 (eq-equation)
x/4 + 4y = 8 --------- eq2
A =
1 16
¼ 4
A/b =
1 16 9
¼ 4 8
Rank of A:
row2  4*row2 - row1
1 16
0 0
 rank of A = non zero rows of A = 1
Rank of A/b:
row2  4*row2 - row1
1 16 9
0 0 23


rank of A/b = non zero rows of A/b = 2
rank of A ≠ rank of A/b
 it has no solution (option D)
4) 3x - 2y = 7 ---------eq1 (eq-equation)
9x - 6y = 14 --------- eq2
A =
3 -2
9 -6
A/b =
3 -2 7
9 -6 14
Rank of A:
row2  row2 - 3*row1
3 -2
0 0
 rank of A = non zero rows of A = 1
Rank of A/b:
row2  row2 - 3*row1
1 16 9
0 0 -7
rank of A/b = non zero rows of A/b = 2
rank of A ≠ rank of A/b
 it has no solution (option D)
5) x + 3y = 9 ---------eq1 (eq-equation)
3x- y = 7 --------- eq2
A =
1 3
3 -1
Det(A) = (1*-1) – (3*3) = -10 ≠ 0
 it has only one solution
eq1 * 3 => 3x + 9y = 27
eq2 => 3x - y = 7
eq1*3 - eq2 = 10 y = 20 => y = 2
substitute y = 2 in eq1 => x + (3* 2) = 9 => x = 3
option A
6) 2x - 5y = 10 ---------eq1 (eq-equation)
4x - 10y = 20 --------- eq2
A =
2 -5
4 -10
A/b =
2 -5 10
4 -10 20
Rank of A:
row2  row2 - 2*row1
2 -5
0 0
 rank of A = non zero rows of A...