Linear Algebra Question of the Day

1. $w = \begin{pmatrix} 1 \\ 1 \\ -2 \\ 0 \end{pmatrix}$
2. $w = \begin{pmatrix} 1 \\ 1 \\ 2 \\ 0 \end{pmatrix}$
3. $w = \begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$
4. $w = \begin{pmatrix} 1 \\ 3 \\ -2 \\ 1 \end{pmatrix}$ (correct answer)

Explanation: The row space of $A$ is the orthogonal complement of the null space of $A$ in $\mathbb{R}^4$. A vector $w$ is in the row space if and only if it is orthogonal to every vector in the basis for the null space. Let the basis vectors be $v_1$ and $v_2$. We need to find the choice $w$ for which $w \cdot v_1 = 0$ and $w \cdot v_2 = 0$. Let's test choice D: $w = (1, 3, -2, 1)$.\n$w \cdot v_1 = (1)(1) + (3)(-1) + (-2)(0) + (1)(2) = 1 - 3 + 0 + 2 = 0$. \n$w \cdot v_2 = (1)(0) + (3)(2) + (-2)(1) + (1)(-1) = 0 + 6 - 2 - 1 = 3$. Oh, calculation error again. Let's fix this. I need a vector orthogonal to both. Let $w=(w_1,w_2,w_3,w_4)$. Then $w_1-w_2+2w_4=0$ and $2w_2+w_3-w_4=0$. Let $w_4=2, w_2=1$. Then $w_3 = w_4-2w_2 = 2-2=0$. And $w_1 = w_2-2w_4 = 1-4=-3$. So $w=(-3, 1, 0, 2)$ is in the row space. Let's make this choice D. \nLet's test the other choices with the original basis vectors. A: $w=(1,1,-2,0)$. $w \cdot v_1 = 1-1=0$. $w \cdot v_2 = 2-2=0$. So A is correct. Let me check my initial setup. Maybe I made a mistake somewhere else. Oh, I see. I must have intended for only one to work. Let me check the other initial choices with the basis. B: $w=(1,1,2,0)$. $w \cdot v_1 = 1-1=0$. $w \cdot v_2 = 2+2=4$. So B is wrong. C: $w=(2,0,1,-1)$. $w \cdot v_1 = 2-2=0$. $w \cdot v_2 = 1+1=2$. So C is wrong. So my choice A was already correct. Let me re-verify D: $w=(1,3,-2,1)$. $w \cdot v_1 = 1-3+2=0$. $w \cdot v_2 = 6-2-1=3$. So D is wrong. The correct answer is A.