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Does there exists a matrix A such that \(\mathbf{A} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 0 \end{pmatrix},\mathbf{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \mathbf{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\)
calculate \(\mathbf{A} \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}. \)
if so, what is it?

 Oct 18, 2020
 #1
avatar+111124 
+1

I do not think that it exists.

Let

A = \(\begin{pmatrix} a\;b\;c\\ d\;e\;f\\ h\;i\;j\\ \end{pmatrix}\)

 

\(\begin{pmatrix} a\;b\;c\\ d\;e\;f\\ h\;i\;j\\ \end{pmatrix} \cdot \begin{pmatrix} 1\\0\\0 \end{pmatrix}= \begin{pmatrix} a\\d\\h \end{pmatrix}= \begin{pmatrix} -3\\4\\0 \end{pmatrix}\\ so\\ a=-3\qquad d=4\qquad h=0\)

 

 

\(\begin{pmatrix} -3\;b\;c\\ 4\;e\;f\\ 0\;i\;j\\ \end{pmatrix} \cdot  \begin{pmatrix} 0\\1\\1 \end{pmatrix}= \begin{pmatrix} b+c\\e+f\\i+j \end{pmatrix}= \begin{pmatrix} 1\\2\\3 \end{pmatrix}\\ so\\ c=1-b \qquad f=2-e\qquad j=3-i\)

 

 

\(\begin{pmatrix} -3\;\;b\;\;1-b\\ 4\;\;e\;\;2-e\\ 0\;i\;\;\;3-i\\ \end{pmatrix} \cdot  \begin{pmatrix} 1\\1\\1 \end{pmatrix}= \begin{pmatrix} -3+1\\4+2\\0+3 \end{pmatrix}= \begin{pmatrix} -2\\6\\3 \end{pmatrix}\ne \begin{pmatrix} 3\\2\\1 \end{pmatrix}\\ \)

 

 

 

 

LaTex:

\text{\begin{pmatrix}
a\;b\;c\\
d\;e\;f\\
h\;i\;j\\
\end{pmatrix}

 

\begin{pmatrix}
a\;b\;c\\
d\;e\;f\\
h\;i\;j\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
1\\0\\0
\end{pmatrix}=
\begin{pmatrix}
a\\d\\h
\end{pmatrix}=
\begin{pmatrix}
-3\\4\\0
\end{pmatrix}\\
so\\ a=-3\qquad b=4\qquad c=0

 

\begin{pmatrix}
-3\;b\;c\\
4\;e\;f\\
0\;i\;j\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
0\\1\\1
\end{pmatrix}=
\begin{pmatrix}
b+c\\e+f\\i+j
\end{pmatrix}=
\begin{pmatrix}
1\\2\\3
\end{pmatrix}\\
so\\ c=1-b \qquad f=2-e\qquad j=3-i

 

\begin{pmatrix}
-3\;\;b\;\;1-b\\
4\;\;e\;\;2-e\\
0\;i\;\;\;3-i\\
\end{pmatrix}
\cdot 
\begin{pmatrix}
1\\1\\1
\end{pmatrix}=
\begin{pmatrix}
-3+1\\4+2\\0+3
\end{pmatrix}=
\begin{pmatrix}
-2\\6\\3
\end{pmatrix}\ne
\begin{pmatrix}
3\\2\\1
\end{pmatrix}\\}

 

 
 Oct 18, 2020
 #2
avatar+2452 
+2

Melody for some reason, your latex or the part below doesn't show up for me sad

 

Is there a reason why?

 
 Oct 18, 2020
 #3
avatar+111124 
+1

Hi Cal,

Yes, it refused to display as code. It displayed as not properly formatted Latex and I did not want that.

So I made the writing white so it would disappear.

I think you can highlight and copy it if anyone wanted to see it.

 

No one has ever asked me why I usually add the coding at the bottom like this. This has surprised me a little.

I do it for 2 reasons.

1) It may help some people to learn LaTex Coding.

2) Web 2 often will not allow me to get in and edit my code, especially not when it is a big block of code.

   If I have the code at the bottom I can copy it into a new Latex box and edit that instead (Then i would delete the original)

 
Melody  Oct 18, 2020
 #4
avatar+2452 
+1

I also think that Latex is good and necessary to learn. And yes, webcalc2.0 sometimes censors Latex.

That makes sense. Thanks for the answer Melody! laugh

 
CalTheGreat  Oct 18, 2020
 #5
avatar+111124 
0

It was not censored.

I wanted to display the code not the output but it refused to display that way.

So I whited  (literally) it out.

 
Melody  Oct 19, 2020
 #6
avatar+2452 
+1

Oh, okay, I get it! 

 

can you read this? :)

 
CalTheGreat  Oct 19, 2020

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