Question

1. Find the solution(s) of the following linear systems. Use the MATLAB command rref to immediately find the reduced row echelon form of a matrix. Enter the coefficient matrix A and vector b. When entering the vector b, type b=[b1 b2 b3]'. The ' indicates transpose and changes b from a row vector to a column vector. Type rref([A b]). This is asking for the reduced row echelon form of the augmented matrix [A b]. For each system manually type [Times New Roman] in all solutions (or state that there is no solution) on your printout near the problem. (a) $2x_1 + x_2 + 3x_3 + 2x_4 = -4$ $-x_1 + 2x_2 + x_3 + x_4 = 2$ $3x_1 - 2x_2 + x_3 + 4x_4 = -21$ $4x_1 + 7x_2 + 5x_3 + 4x_4 = 12$ (b) $2x_1 + 6x_2 - 3x_3 - x_4 + 18x_5 + 2x_6 = -103$ $x_2 + 3x_3 + 4x_4 + 12x_5 + 9x_6 = -61$ $-2x_1 + 3x_2 + 8x_3 + 12x_4 - 5x_5 + 6x_6 = 63$ $-7x_1 + 4x_2 - x_3 + 5x_4 + 39x_5 + x_6 = -208$ $7x_1 + 4x_2 - x_3 + 5x_4 + 66x_5 = -469$ (c) $2x_1 + 6x_2 - 3x_3 = -1$ $-2x_1 + x_2 + x_3 = 4$ $-x_1 + 7x_2 + 8x_3 = 12$ $7x_1 + 4x_2 - 3x_3 = 5$ (d) $x_1 - x_2 + x_3 - x_4 = 0$ $x_1 + 2x_2 - 3x_3 + 4x_4 = 0$ $3x_1 - x_2 - 2x_3 + 2x_4 = 0$

          1. Find the solution(s) of the following linear systems. Use the MATLAB command rref to immediately find the reduced row echelon form of a matrix. Enter the coefficient matrix A and vector b. When entering the vector b, type b=[b1 b2 b3]'. The ' indicates transpose and changes b from a row vector to a column vector. Type rref([A b]). This is asking for the reduced row echelon form of the augmented matrix [A b]. For each system manually type [Times New Roman] in all solutions (or state that there is no solution) on your printout near the problem.
(a)
$2x_1 + x_2 + 3x_3 + 2x_4 = -4$
$-x_1 + 2x_2 + x_3 + x_4 = 2$
$3x_1 - 2x_2 + x_3 + 4x_4 = -21$
$4x_1 + 7x_2 + 5x_3 + 4x_4 = 12$
(b)
$2x_1 + 6x_2 - 3x_3 - x_4 + 18x_5 + 2x_6 = -103$
$x_2 + 3x_3 + 4x_4 + 12x_5 + 9x_6 = -61$
$-2x_1 + 3x_2 + 8x_3 + 12x_4 - 5x_5 + 6x_6 = 63$
$-7x_1 + 4x_2 - x_3 + 5x_4 + 39x_5 + x_6 = -208$
$7x_1 + 4x_2 - x_3 + 5x_4 + 66x_5 = -469$
(c)
$2x_1 + 6x_2 - 3x_3 = -1$
$-2x_1 + x_2 + x_3 = 4$
$-x_1 + 7x_2 + 8x_3 = 12$
$7x_1 + 4x_2 - 3x_3 = 5$
(d)
$x_1 - x_2 + x_3 - x_4 = 0$
$x_1 + 2x_2 - 3x_3 + 4x_4 = 0$
$3x_1 - x_2 - 2x_3 + 2x_4 = 0$

1. Find the solution(s) of the following linear systems. Use the MATLAB command rref to immediately find the reduced row echelon form of a matrix. Enter the coefficient matrix A and vector b. When entering the vector b, type b=[b1 b2 b3]'. The ' indicates transpose and changes b from a row vector to a column vector. Type rref([A b]). This is asking for the reduced row echelon form of the augmented matrix [A b]. For each system manually type [Times New Roman] in all solutions (or state that there is no solution) on your printout near the problem.
(a)
2x1 + x2 + 3x3 + 2x4 = -4
-x1 + 2x2 + x3 + x4 = 2
3x1 - 2x2 + x3 + 4x4 = -21
4x1 + 7x2 + 5x3 + 4x4 = 12
(b)
2x1 + 6x2 - 3x3 - x4 + 18x5 + 2x6 = -103
x2 + 3x3 + 4x4 + 12x5 + 9x6 = -61
-2x1 + 3x2 + 8x3 + 12x4 - 5x5 + 6x6 = 63
-7x1 + 4x2 - x3 + 5x4 + 39x5 + x6 = -208
7x1 + 4x2 - x3 + 5x4 + 66x5 = -469
(c)
2x1 + 6x2 - 3x3 = -1
-2x1 + x2 + x3 = 4
-x1 + 7x2 + 8x3 = 12
7x1 + 4x2 - 3x3 = 5
(d)
x1 - x2 + x3 - x4 = 0
x1 + 2x2 - 3x3 + 4x4 = 0
3x1 - x2 - 2x3 + 2x4 = 0

Added by Ryan A.

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