307w.43.M40 find corresponding 3 scalars

[karush]

ok did an image due to notation to be correct, we will be talking about this in the zoom class next week
but wanted to some grip of it before. Here is the RREFs
$\text{rref}(B)=\left[ \begin{array}{cccc} 1 & 0 & 7 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$
I assume we can switch c3 and c4 to get the triangle
and

$\text{rref}(C)=\left[ \begin{array}{cccc} 1 & 0 & 7 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$

 

[HOI]

That looks good. You started with $B= \begin{bmatrix}1 & 3 & -2 & 2 \\ -1 & -2 & -1 & -1 \\ -1 & -5 & 8 & -3\end{bmatrix}$ and, I presume, added the first row to both second and third rows to get $\begin{bmatrix}1 & 3 & -2 & 2 \\ 0 & 1 & -3 & 1 \\ 0 & -2 & 6 & -1\end{bmatrix}$ and then add twice the second row to the third to get $\begin{bmatrix}1 & 3 & -2 & 2 \\ 0 & 1 & -3 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$. Now subtract three times the new second row from the first row to get $\begin{bmatrix}1 & 0 & 7 & -1 \\ 0 & 1 & -3 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$ and, finally, add the new third row to the first row and subtract the new third row from the second row to get $\begin{bmatrix}1 & 0 & 7 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$

And with $C= \begin{bmatrix}1 & 2 & 1 & 2 \\ 1 & 1 & 4 & 0 \\ -1 & -1 & -4 & 1 \end{bmatrix}$ I think I would first add the second row to the third, then subtract the first row from the second to get $\begin{bmatrix}1 & 2 & 1 & 2 \\ 0 & -1 & 3 & -2 \\ 0 & 0 & 0 & 1 \end{bmatrix}$. Subtract twice the third row from the first row and add twice the third row to the second row to get $\begin{bmatrix}1 & 2 & 1 & 0 \\ 0 & -1 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$. Finally, add twice the new second row to the first row, then multiply the second row by -1 to get $\begin{bmatrix}1 & 0 & 7 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$. Yes, the two matrices row-reduce to the same thing.

Now to do the rest of the exercise, write each row of B as a linear combination of rows of C, you can use each of those row-reductions. Each row of that row reduction is a linear combination of rows of C. For example the only change in the first row of C was when we added "twice the new second row to the first row" and the "new second row" was formed when we added "twice the third row to the second row". Writing the first, second, and third rows as $c_1$, $c_2$, and $c_2$, respectively, adding twice the third row to the second row gives $c_2+ 2c_3$ and adding twice that to first row gives $c_1+ 2(c_2+ 2c_3)= c_1+ cr_2+ 4c_3$. Do that for the second and third rows to get the rows of the common row-reduced matrix as linear combinations of the rows of C.

Do the same to get the rows of the common row reduced matrix in term of the rows of B, $b_1$, $b_2$, and $b_3$. Then set those equal to the expressions in terms of $c_1$, $c_2$, and $c_3$ and solve for $b_1$, $b_2$, and $b_3$ in terms of $c_1$, $c_2$, and $c_3$!