Answer
See belowGiven: $A=\begin{bmatrix}
1 & 3 & -2 & 1\\3 & 10 & -4 & 6\\ 2 & 5 & -6 & -1
\end{bmatrix}$
Obtain: $\begin{bmatrix}
1 & 3 & -2 & 1\\3 & 10 & -4 & 6\\ 2 & 5 & -6 & -1
\end{bmatrix}\begin{bmatrix}
x_1\\x_2\\x_3\\ x_4
\end{bmatrix}=0$
We have the system:
$x_1+3x_2−2x_3+x_4=0\\3x_1+10x_2−4x_3+6x_4=0\\2x_1+5x_2−6x_3−1x_4=0$
From exercise 27, we have $x_1=8x_3+8x_4\\x_2=-2x_3-3x_4$
and $ x_1=(8x_3+8x_4,−2x_4−3x_4,x_3,x_4)\\=(8x_3,−2x_3,x_3,0)+(8x_4,−3x_4,0,x_4)\\=x_3(8,−2,1,0)+x_4(8,−3,0,1)$
Then $N(A)=((8,−2,1,0),(8,−3,0,1))$
Hence, $N(A)=span (8,−2,1,0),(8,−3,0,1)$
Work Step by Step
Given: $A=\begin{bmatrix}
1 & i & -2 \\3 & 4i & -5\\ -1 & -3i & i
\end{bmatrix}$
From exercise 29, we have $N(A)=(0,0,0)$
Hence, $N(A)=span (0,0,0)$
