Practice Test Solutions: Derivation of [ -4/3, 1/3, 1 ] with Red Correction

[karush]

View attachment 8987

Ok I didn't understand how he derived the

$\left[\begin{array}{c}-4/3\\1/3\\1 \end{array}\right]$

where the red correction is

 

[Ackbach]

For the rref form of $A,$ I get
$$A_{\text{rref}}=\left[\begin{matrix}1&0&4/3\\0&1&-1/3\\0&0&0\end{matrix}\right].$$
To find the null space of $A,$ you're essentially solving $A\mathbf{x}=0$ for $\mathbf{x}$. You can see that you have the equations
\begin{align*}
x+(4/3)z&=0\\
y-(1/3)z&=0
\end{align*}
to solve. One solution is $z=1, y=1/3, x=-4/3.$ Any multiple of this vector is also a solution, as $A(t\mathbf{x})=t(A\mathbf{x})=t\cdot 0=0.$

I'm not entirely sure the author's form of $A_{\text{rref}}$ is correct, but the null space calculation is correct.

 

[karush]

ok couldn't see where the negative came from

 

[HallsofIvy]

Personally, I wouldn't have reduced the matrix. The null space is, of course, the set of all $$\begin{bmatrix}x \\ y \\ z \end{bmatrix}$$ such that $$\begin{bmatrix}1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 2 & -2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix} x+ y+ z \\ 2x- y+ 3z \\ -x+ 2y- 2z\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ which is the same as the three equations
x+ y+ z= 0
2x- y+ 3z= 0
-x+ 2y- 2z= 0.

Adding the first two equations eliminates y leaving 3x+ 4z= 0. Adding twice the second equation to the third also eliminates y and leaves the same 3x+ 4z. Any $$\begin{bmatrix}x \\ y \\ z \end{bmatrix}$$ in the null space must satisfy 3x+ 4z= 0. Solve that for x: $$x= -\frac{4}{3}z$$. Then $$x+ y+ z= -\frac{4}{3}z+ y+ z= y- \frac{1}{3}z= 0$$ so $$y= \frac{1}{3}z$$s. Any vector in the null space is of the form $$\begin{bmatrix}-\frac{4}{3} \\ \frac{1}{3} \\ 1\end{bmatrix}z$$. We can take z= 1 and say that any vector in the null space is a multiple of $$\begin{bmatrix}-\frac{4}{3}\\ \frac{1}{3} \\ 1\end{bmatrix}$$.

(Some of use, who don't like fractions, might be inclined to take z= -3 and say that any vector in the null space is a multiple of $$\begin{bmatrix}4 \\ -1 \\ -3 \end{bmatrix}$$.)