Question

Problem No. 2.4 0PS 2.xi-7x2-5x3 1 -7x1-32-9x3-3 Solve the system of linear equations by modifying it to REF and to RREF using elementary equivalent operations. Show REF and RREF of the system Show all your work, do not skip steps Displaying only answer is not enough to get credit Matrices may not be used

Answer 1

The augmented matrix for the given set of equations is

The augmented matrix for the given set of equations is [2 -4 -7 -7 3 -3 -5 -5 -9 1 -4 -3] Row Echelon Form (REF) R_1 rightarrow R_1/2 [1 -4 -7 -7/2 3 -3 -5/2 -5 -9 1/2 -4 -3] R_2 rightarrow 4R_1 + R_2 [1 0 -7 -7/2 -11 -3 -5/2 -15 -9 1/2 -2 -3] R_3 rightarrow 7R_1 + R_3 [1 0 0 -7/2 -11 -55/2 -5/2 -15 -53/2 1/2 -2 1/2] R_2 rightarrow -R_2/11 [1 0 0 -7/2 1 -55/2 -5/2 15/11 -53/2 1/2 2/11 1/2] R_3 rightarrow 55R_2/2 + R_3 [1 0 0 -7/2 1 0 -5/2 15/11 11 1/2 2/11 11/2] R_3 rightarrow R_3/11 [1 0 0 -7/2 1 0 -5/2 15/11 1 1/2 2/11 1/2] The transformed equation from the above REF matrix are x_3 = 1/2 x_2 + 15x_3/11 = 2/11 = > x_2 = 2/11 - 15/11(1/2) = 4/22 - 15/22 = -11/22 = -1/2 x_1 - 7x_2/2 - 5x_3/2 = 1/2 = > x_1 = 1/2

(a)

Row Echelon Form (REF)

The transformed equation from the above REF matrix are

$x_3=1/2$

$x_2+15x_3/11=2/11$

$r1-7x212-513/2=1/2$

So the solution is

(b)

Reduced Row Echelon Form (RREF)

From the above matrix, the solution to the given set of equations is