题目

线性方程组[}x_1 + x_2 + x_3 + x_4 + x_5 = 1 3x_1 + 2x_2 + x_3 + x_4 - 3x_5 = c x_2 + 2x_3 + 2x_4 + 6x_5 = 3 5x_1 + 4x_2 + 3x_3 + 3x_4 - x_5 = d]当 (1) c = ()，(2) d = () 时，有解有解时，求得通解[( x_1 x_2 x_3 x_4 x_5 )]则 (1) = ( )、(2) = ( )、(3) = ( )、(4) = ( )、(5) = ( )、(6) = ( )、(7) = ( )、(8) = ( )、(9) = ( )、(10) = ( )、(11) = ( )、(12) = ( )、(13) = ( )、(14) = ( )、(15) = ( )、(16) = ( )、(17) = ( )、(18) = ( )、(19) = ( )、(20) = ( )、(21) = ( )、(22) = ( )

线性方程组 $\begin{cases} x_1 + x_2 + x_3 + x_4 + x_5 = 1 \\ 3x_1 + 2x_2 + x_3 + x_4 - 3x_5 = c \\ x_2 + 2x_3 + 2x_4 + 6x_5 = 3 \\ 5x_1 + 4x_2 + 3x_3 + 3x_4 - x_5 = d \end{cases}$ 当 (1) $c = ()$，(2) $d = ()$ 时，有解有解时，求得通解 $\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array} \right) = c_1 \left( \begin{array}{c} (3) \\ (4) \\ (5) \\ (6) \\ (7) \end{array} \right) + c_2 \left( \begin{array}{c} (8) \\ (9) \\ (10) \\ (11) \\ (12) \end{array} \right) + c_3 \left( \begin{array}{c} (13) \\ (14) \\ (15) \\ (16) \\ (17) \end{array} \right) + \left( \begin{array}{c} (18) \\ (19) \\ (20) \\ (21) \\ (22) \end{array} \right)$ 则 (1) = ( )、(2) = ( )、(3) = ( )、(4) = ( )、(5) = ( )、(6) = ( )、(7) = ( )、(8) = ( )、(9) = ( )、(10) = ( )、(11) = ( )、(12) = ( )、(13) = ( )、(14) = ( )、(15) = ( )、(16) = ( )、(17) = ( )、(18) = ( )、(19) = ( )、(20) = ( )、(21) = ( )、(22) = ( )

题目解答

答案

当 $c = 0$ 且 $d = 2$ 时，方程组有解。通解为： \[ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ -2 \\ 0 \\ 1 \\ 0 \end{pmatrix} + c_3 \begin{pmatrix} 5 \\ -6 \\ 0 \\ 0 \\ 1 \end{pmatrix} + \begin{pmatrix} -2 \\ 3 \\ 0 \\ 0 \\ 0 \end{pmatrix} \] 因此，答案为： \[ (1) = 0, \quad (2) = 2 \] \[ (3) = 1, \quad (4) = -2, \quad (5) = 1, \quad (6) = 0, \quad (7) = 0 \] \[ (8) = 1, \quad (9) = -2, \quad (10) = 0, \quad (11) = 1, \quad (12) = 0 \] \[ (13) = 5, \quad (14) = -6, \quad (15) = 0, \quad (16) = 0, \quad (17) = 1 \] \[ (18) = -2, \quad (19) = 3, \quad (20) = 0, \quad (21) = 0, \quad (22) = 0 \]