null space of matrix with unknown value

Question

anna_1313 · Answer

attachment

anna_1313 · Answer

i know how to compute the null space, but I'm not sure how to do it when there is an unknown value t in the matrix, it says the answer, could someone please explain it to me?

holsteremission · Answer

Suppose you write $\mathbf{A}$ in row-reduced echelon form $\mathbf{R}$:
$$\begin{align*}
\mathbf{A}=\begin{bmatrix}1&t&0&1\-1&1&0&1\1&-1&0&1\end{bmatrix}&\sim\begin{bmatrix}1&t&0&1\0&1+t&0&2\0&-1-t&0&0\end{bmatrix}\[1ex]
&\sim\begin{bmatrix}1&t&0&1\0&1+t&0&2\0&0&0&2\end{bmatrix}\[1ex]
&\sim\begin{bmatrix}1&0&0&0\0&1&0&0\0&0&0&1\end{bmatrix}=\mathbf{R}
\end{align*}$$which suggests that $\begin{bmatrix}0&0&c&0\end{bmatrix}^\intercal$ is in the nullspace (provided that $t\neq-1$, as the hint suggests). In order for $\mathbf{R}\mathbf{x}=\mathbf{0}$ to be true, the first, second and fourth components must be zero, while there is no such constraint on the third component.

holsteremission · Answer

Actually, I misunderstood the hint - the vector $\begin{bmatrix}0&0&c&0\end{bmatrix}^\intercal$ is in the nullspace regardless of the value of $t$.

anna_1313 · Answer

how did 1+t dissapear in the 3rd matrix become 1 in the 4th matrix?

holsteremission · Answer

Whoops, I must have skipped a few steps there. The row operations I used were (in order)

1. $R_2+R_1\to R_1$ and $R_3-R_1\to R_3$ simultaneously
2. $R_3+R_2\to R_3$

Not shown are

3. $R_2-R_3\to R_2$, which returns
$$\begin{bmatrix}1&t&0&1\0&1+t&0&2\0&0&0&2\end{bmatrix}\to\begin{bmatrix}1&t&0&1\0&1+t&0&0\0&0&0&2\end{bmatrix}$$4. $-\dfrac{1}{1+t}R_2\to R_2$ (which can only be done if $t\neq-1$) and $-\dfrac{1}{2}R_3\to R_3$ simultaneously, which gives
$$\begin{bmatrix}1&t&0&1\0&1+t&0&0\0&0&0&2\end{bmatrix}\to\begin{bmatrix}1&t&0&1\0&1&0&0\0&0&0&1\end{bmatrix}$$5. $R_1-tR_2-R_3\to R_1$, which gives the rref matrix $\mathbf{R}$.