Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.3

Solution

A homogeneous system has the form AX=0AX=0.


(i)

{x+4y2z=02x+y+5z=05x+2y+8z=0\begin{cases} x+4y-2z=0\\ 2x+y+5z=0\\ 5x+2y+8z=0 \end{cases}

Row-reduction gives rank 33, so only the trivial solution exists:

x=0,  y=0,  z=0\boxed{x=0,\; y=0,\; z=0}

(ii)

{x1+4x2+2x3=02x1+x23x3=03x1+2x24x3=0\begin{cases} x_1+4x_2+2x_3=0\\ 2x_1+x_2-3x_3=0\\ 3x_1+2x_2-4x_3=0 \end{cases}

RREF gives:

x12x3=0x2+x3=0\begin{aligned} x_1-2x_3&=0\\ x_2+x_3&=0 \end{aligned}

Let x3=tx_3=t. Then x1=2tx_1=2t and x2=tx_2=-t.

(x1,x2,x3)=t(2,1,1)\boxed{(x_1,x_2,x_3)=t(2,-1,1)}

(iii)

{x1+2x2x3=0x1x2+5x3=02x1+x2+4x3=0\begin{cases} x_1+2x_2-x_3=0\\ x_1-x_2+5x_3=0\\ 2x_1+x_2+4x_3=0 \end{cases}

RREF gives:

x1+3x3=0x22x3=0\begin{aligned} x_1+3x_3&=0\\ x_2-2x_3&=0 \end{aligned}

Let x3=tx_3=t. Then x1=3tx_1=-3t and x2=2tx_2=2t.

(x1,x2,x3)=t(3,2,1)\boxed{(x_1,x_2,x_3)=t(-3,2,1)}