Linear algebra help please..... see attachment

Question

chrisplusian · Answer

attachment

chrisplusian · Answer

I need help with 8 and 9

thomas5267 · Answer

@ganeshie8  @IrishBoy123 I am not sure about this.

I would say $\operatorname{rank}(\mathbf{A})=1$ since the vector corresponding to $\alpha$ and $\beta$ must lie in the null space of $\mathbf{A}$.

I have no idea what is 8(i) and 8(iii) is asking though.

For 9(iii), the rank is 3?  Suppose the $\operatorname{rank}(\mathbf{A})<3$ and $\mathbf{Ax}=\begin{pmatrix}1&3&5&7&9\end{pmatrix}^T$ still holds ($\mathbf{x}=\begin{pmatrix}0&-1&0\end{pmatrix}^T$).  Since $\operatorname{rank}(\mathbf{A})<3$, there exist at least one vector $\mathbf{v}$ where $\mathbf{Av}=\begin{pmatrix}0&0&0&0&0\end{pmatrix}^T$.  But $\mathbf{A}(\mathbf{x}+\alpha \mathbf{v})=\mathbf{Ax}+\alpha\mathbf{Av}=\begin{pmatrix}1&3&5&7&9\end{pmatrix}^T$ since $\mathbf{Av}=\begin{pmatrix}0&0&0&0&0\end{pmatrix}^T$ by definition.  Since we can choose $\alpha$ as we please, there are infinite amount of solutions so $\operatorname{rank}(\mathbf{A})=3$.

ganeshie8 · Answer

8(i) 

I think it should be
$$R=\begin{bmatrix} 1&-2&-5\0&0&0\0&0&0\end{bmatrix}$$

8(ii) rank = $1$

8(iii)
$$d=\begin{pmatrix}4\0\0\end{pmatrix}$$

thomas5267 · Answer

That makes sense.  Since it is a rank 1 matrix, it must have two 0 rows.  Since it is in reduced row echelon form, the two zero rows must be the last two rows.  Since the two vectors must be in the null space, so A(that two vectors) must be 0.  So the matrix is:
$$
\begin{pmatrix}
c_1&c_2&c_3\
0&0&0\
0&0&0
\end{pmatrix}
$$

chrisplusian · Answer

@ganeshie8 can you explain? I agree with what @thomas5267 said. I just don't understand how exactly  you come up with the numeric values you did.

empty · Answer

I was looking at x, and saw this:

$$
\begin{pmatrix}
4 &2  &5 \ 
0 &1  &0 \0 
 &0  &1 
\end{pmatrix} \begin{pmatrix}
1\ \alpha 
\ \beta 

\end{pmatrix} = \bar x $$

Which appears to be $$A^{-1} \bar b = \bar x$$ if you left multiply by A, you get:

$$\bar b = A \bar x$$ 

But I think I am assuming too much here in saying that this matrix and vector are A inverse and b.  But they might be related by a similarity transformation... which is probably overthinking this problem, idk. Food for thought I guess.

thomas5267 · Answer

For question 8(i).
$$
\mathbf{R}=\begin{pmatrix}
c_1&c_2&c_3\
0&0&0\
0&0&0
\end{pmatrix}\text{ as }\operatorname{rank}(\mathbf{A})=1\
\mathbf{v_1}=\begin{pmatrix}2\1\0\end{pmatrix}\
\mathbf{v_2}=\begin{pmatrix}5\0\1\end{pmatrix}
$$
Solve $\mathbf{Rv_1}=\mathbf{0},\mathbf{Rv_2}=\mathbf{0}$ for $c_1,c_2,c_3$.  You should get what ganeshie8 got.  Then multiply $\begin{pmatrix}4&0&0\end{pmatrix}^T$ by $\mathbf{R}$ to get $\mathbf{d}$ for 8(iii).