-la.1.1.19 the augmented process

[karush]

nmh{1000} ov 346.21

ok I am trying to these 4 problems
but don't think I understand the augmented process

these are the answers to 19 and 21



$\tiny{1.1.19}$
$$A_{19}=\left[\begin{array}{rrrrr}
1& \,h& \,4\\
3& \, 6& \, 8
\end{array}\right]$$$\textit{so if we multiply $R_1$ by subtract from R_3 we have}$\\
$$\left[\begin{array}{rrrrr}
1& \,h& \,4\\
0& \, (6-3h)& \, -4
\end{array}\right]$$uhmmm ?

 

[I like Serena]

$\tiny{1.1.19}$
$$A_{19}=\left[\begin{array}{rrrrr}
1& \,h& \,4\\
3& \, 6& \, 8
\end{array}\right]$$$\textit{so if we multiply $R_1$ by subtract from R_3 we have}$
$$\left[\begin{array}{rrrrr}
1& \,h& \,4\\
0& \, (6-3h)& \, -4
\end{array}\right]$$uhmmm ?

Good. Can you continue?
What would $y$ become?

 

[karush]

Good. Can you continue?
What would $y$ become?

with (h=y)
\begin{align*}
(6-3h)&=-4 \\
3h&=10\\
h&=\frac{10}{3}
\end{align*}but how does this answer $h \ne 2$ ?

 

[I like Serena]

with (h=y)
\begin{align*}
(6-3h)&=-4 \\
3h&=10\\
h&=\frac{10}{3}
\end{align*}

but how does this answer $h \ne 2$ ?

The second line in the augmented matrix actually stands for:
$$0\cdot x + (6-3h)\cdot y = -4 \quad\Rightarrow\quad (6-3h)y=-4$$
Can we solve $y$ from it?

 

[karush]

Re: la.1.1.19 the augumented process

The second line in the augmented matrix actually stands for:
$$0\cdot x + (6-3h)\cdot y = -4 \quad\Rightarrow\quad (6-3h)y=-4$$
Can we solve $y$ from it?

$\begin{align*}
0\cdot x + (6-3h)\cdot y &= -4\\
(6-3h)\cdot y&=-4\\
y&=-\frac{4}{6-3h}
\end{align*}$
so then h equals all real numbers except 2
whatever the notation is for that?

 

[I like Serena]

Re: la.1.1.19 the augumented process

$\begin{align*}
0\cdot x + (6-3h)\cdot y &= -4\\
(6-3h)\cdot y&=-4\\
y&=-\frac{4}{6-3h}
\end{align*}$
so then h equals all real numbers except 2
whatever the notation is for that?

Good. It means that $y$ has a solution if and only if $h\ne 2$.

Btw, for the system to be linearly consistent, we also need to be able to find a solution for $x$.
Can we?

 

[karush]

Re: la.1.1.19 the augumented process

Good. It means that $y$ has a solution if and only if $h\ne 2$.

Btw, for the system to be linearly consistent, we also need to be able to find a solution for $x$.
Can we?

I can only see that x could be any real number since it would be gone by multiplying by zero

ok here is the last one but only see a single answer?

$A_{22}=\left[\begin{array}{rrrrr}
-4& \,12& \,\textbf{h}\\ 2& -6& -3 \end{array}\right]\\$
$\text{Add $R_2$ to $R_1$}\\$
$\left[\begin{array}{rrrrr} -4+2& \,12-6& \, h-3\\ 2& -6& -3\end{array}\right]\\
0+0=h-3\\$
$\text{so $h=3$}$

 

[I like Serena]

Re: la.1.1.19 the augumented process

I can only see that x could be any real number since it would be gone by multiplying by zero

What about the first line of the augmented matrix?
Can't we deduce $x$ from it now that we have an expression for $y$?

ok here is the last one but only see a single answer?

$A_{22}=\left[\begin{array}{rrrrr}
-4& \,12& \,\textbf{h}\\ 2& -6& -3 \end{array}\right]\\$
$\text{Add $R_2$ to $R_1$}\\$
$\left[\begin{array}{rrrrr} -4+2& \,12-6& \, h-3\\ 2& -6& -3\end{array}\right]\\
0+0=h-3\\$
$\text{so $h=3$}$

How about bringing it into row echelon form as before?
And solve for $y$ again?

 

[karush]

Re: la.1.1.19 the augumented process

$\left[\begin{array}{rrrrr}
-4& \,12& \,\textbf{h}\\ 2& -6& -3 \end{array}\right]\\$
Ok to avoid fraction switch $R_1$ and $R_2$
$\left[\begin{array}{rrrrr} 2& -6& -3 \\ -4& \,12& \,\textbf{h} \end{array}\right]\\$
then multiply R_1 by 2 and add it to R_2
$\left[\begin{array}{rrrrr} 2& -6& -3 \\ 0& \,0 & \,\textbf{h-6} \end{array}\right]\\$
then
0=h-6
so
h=6

 

[I like Serena]

Re: la.1.1.19 the augumented process

$\left[\begin{array}{rrrrr}
-4& \,12& \,\textbf{h}\\ 2& -6& -3 \end{array}\right]\\$
Ok to avoid fraction switch $R_1$ and $R_2$
$\left[\begin{array}{rrrrr} 2& -6& -3 \\ -4& \,12& \,\textbf{h} \end{array}\right]\\$
then multiply R_1 by 2 and add it to R_2
$\left[\begin{array}{rrrrr} 2& -6& -3 \\ 0& \,0 & \,\textbf{h-6} \end{array}\right]\\$
then
0=h-6
so
h=6

Correct.
In this case we don't really get to the point that we can solve for $y$.
We can already see that $h$ must be $6$ because otherwise there can't be a solution.

 

[karush]

https://dl.orangedox.com/wlKD7eKSWiQ79alYD6