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Linear Algebra - finding matrix A

  • #1

Homework Statement


Find A if (2A-1 - 3I)T =

2*[tex]
\begin{pmatrix}
-1 & 2\\
5 & 4
\end{pmatrix}
[/tex]

Homework Equations





The Attempt at a Solution



I have no idea if I'm even on the right track of solving this question...

I simplified the right hand side down to
[tex]
\begin{pmatrix}
-2 & 4\\
10 & 8
\end{pmatrix}
[/tex]

I took the transpose of both sides so the left hand side is just (2A-1 - 3I)
The right hand side is now
[tex]
\begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}
[/tex]

I multiplied both sides by A so
LH: A(2A-1 - 3I)
RH: A*[tex]
\begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}
[/tex]

Distribute the A so
LH: (A2A-1 - A3I) => (2I - A3I)

Bring the A3I part to the right side
LH = 2I
RH = A3I + A*[tex]
\begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}
[/tex]

Simplify the A3I part on the right hand side to get
A * [tex]
\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix}
[/tex]

Factor out the A on the right hand side to get
A * ([tex]
\begin{pmatrix}
-2 & 10\\
4 & 8
\end{pmatrix}
[/tex]

+
[tex]
\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix}
[/tex]
)

Add the two matrices together in the brackets to get
A * [tex]\begin{pmatrix}
1 & 10\\
4 & 11
\end{pmatrix}
[/tex]

So I'm left with
[tex]
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}
[/tex]

=

A *
[tex]
\begin{pmatrix}
1 & 10\\
4 & 11
\end{pmatrix}
[/tex]

Now there's no way I can somehow get A like that so I screwed up somewhere in there.. probably from the start lol

This looks so messy... am I allowed to upload my handwritten work on a site then post it here? Seems like a better alternative to the mess I have up there >_> Anyways I'm not really looking for a step by step on how to do this. Just a push in the right direction is all...
 

Answers and Replies

  • #2
33,104
4,800

Homework Statement


Find A if (2A-1 - 3I)T =

2*[tex]
\begin{pmatrix}
-1 & 2\\
5 & 4
\end{pmatrix}
[/tex]

Homework Equations





The Attempt at a Solution



I have no idea if I'm even on the right track of solving this question...

I simplified the right hand side down to
[tex]
\begin{pmatrix}
-2 & 4\\
10 & 8
\end{pmatrix}
[/tex]

I took the transpose of both sides so the left hand side is just (2A-1 - 3I)
The right hand side is now
[tex]
\begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}
[/tex]
It looks like you did a lot more work than you needed to.

You have
$$ 2A^{-1} - 3I = \begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}$$

Just add 3I to both sides, and then you'll have an expression for 2A-1, from which you can fairly easily find A.
I multiplied both sides by A so
LH: A(2A-1 - 3I)
RH: A*[tex]
\begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}
[/tex]

Distribute the A so
LH: (A2A-1 - A3I) => (2I - A3I)

Bring the A3I part to the right side
LH = 2I
RH = A3I + A*[tex]
\begin{pmatrix}
-2& 10\\
4 & 8
\end{pmatrix}
[/tex]

Simplify the A3I part on the right hand side to get
A * [tex]
\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix}
[/tex]

Factor out the A on the right hand side to get
A * ([tex]
\begin{pmatrix}
-2 & 10\\
4 & 8
\end{pmatrix}
[/tex]

+
[tex]
\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix}
[/tex]
)

Add the two matrices together in the brackets to get
A * [tex]\begin{pmatrix}
1 & 10\\
4 & 11
\end{pmatrix}
[/tex]

So I'm left with
[tex]
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}
[/tex]

=

A *
[tex]
\begin{pmatrix}
1 & 10\\
4 & 11
\end{pmatrix}
[/tex]

Now there's no way I can somehow get A like that so I screwed up somewhere in there.. probably from the start lol

This looks so messy... am I allowed to upload my handwritten work on a site then post it here? Seems like a better alternative to the mess I have up there >_> Anyways I'm not really looking for a step by step on how to do this. Just a push in the right direction is all...
 

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