can anyone help me with these two questions please and show me the stepes thanks

Question

can anyone help me with these two questions please and show me the stepes

thanks

Solve the following problems, showing any necessary work. This includes row operations 1. [1 point] How many solutions does each of the following systems of linear equations have? If there are no solutions, circle the row that indicates that there are no solutions. If there is exactly one solution, circle the pivots. Otherwise, circle the columns which do not have pivots in them 0 3 3 12 a.. 0 0 0 2-6 0 0 0 0 L 0 0 0 0 0 0 0 0 -2 -1 b. 0 0 0 0 0 1 0 0 0 0 0 1 2 0 -1 -21 0 0 1 2 2 3 c. 0 0 0 1 -1 0 0 0 0 0 1 -2 0 0 0 0 0 0 L 0 0 0 0 0 2. 12 points] Perform Gauss-Jordan Elimination on the following system of linear equations, and parame- terize the solutions.

Explanation / Answer

(a)The given matrix is an augmented matrix for a linear system of equations in 5 variables , say x1,x2,x3,x4 and x5. The solution is x5 = -2, x4 + 2x5 = -6 ( so that x4 = -2), x2 +3x3 +3x4 -3x5 = 12 ( if x3 = t, then x2 = 12-3t) and x1 –x2 -2x3 -2x4 -2x5 = 1 so that x1 = 1 + x2 +2x3 +2x4 +2x5 = 1 + 12-3t +2t4-4 = 5-t. Thus, there are infinite solutions.

(b) The given matrix is an augmented matrix for a linear system of equations in 5 variables , say x1,x2,x3,x4 and x5. There is no soution and the system is inconsistent. This can be observerd from the 3rd row of the augmented matrix ( 0 cannot be equal to 1).

(c) The given matrix is an augmented matrix for a linear system of equations in 5 variables , say x1,x2,x3,x4 and x5. The solution is x5 = -2, x4 –x5 = 0 (so that x4 = x5 = -2), x3 +2x4+2x5 = 3 (so that x3 = 3 +4+4 = 11), x2 -2x3-2x4-3x5 = -3 ( so that x2 = -3 +22 – 4-6 = 9) and x1 +2x2 –x4 +3x5 = -2 (so that x1 = -2 -18-2 +6 = -16). Thus, there is a unique solution.

2. The augmented matrix of the given linear system is A =

     We will reduce A to its RREF as under:

    Multiply the 1st row by -1

Add 1 times the 1st row to the 2nd row

    Add 3 times the 1st row to the 3rd row

    Then the RREF of A is

Hence the given linear system is equivalent to x4 = 1 and x1 +3x2 -4x3 = 2. Let x2 = r and x3 = t.   Then x1 = 2-3r+4t so that X = (x1,x2,x3,x4)T = (2-3r+4t , r,t,1)T = (2,0,0,1)T + r( -3,1,0,0)T +                       t( 4,0,1,0)T , where r, t are arbitrary real numbers.

-1 -3 4 0 -2 -1 -3 4 1 -1 -3 -9 12 0 -6

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